Let's calculate the interquartile range (IQR).
Given dataset: {5, 7, 9, 12, 15, 18}
1: Order the dataset (it is already ordered in this case): {5, 7, 9, 12, 15, 18}
2: Determine Q1 and Q3. Since there are 6 data points, we will use the following approach to find Q1 and Q3:
Q1: Median of the lower half (including the middle value if the dataset has an odd number of data points)
Q3: Median of the upper half (including the middle value if the dataset has an odd number of data points)
In this case, the dataset has an even number of data points, so we'll find the median of the lower half {5, 7, 9} and the median of the upper half {12, 15, 18}.
Q1: Median of {5, 7, 9} = 7
Q3: Median of {12, 15, 18} = 15
3: Calculate IQR. IQR = Q3 - Q1 = 15 - 7 = 8

To find the variance and standard deviation of a dataset, follow these steps:
Calculate the mean (average) of the dataset.
Subtract the mean from each data point, and square the result.
Calculate the mean of these squared differences.
For the standard deviation, take the square root of the variance.
Let's calculate the variance and standard deviation for the given dataset: 2, 4, 6, 8, 10
Calculate the mean:
(2 + 4 + 6 + 8 + 10) / 5 = 30 / 5 = 6
Subtract the mean from each data point, and square the result:
(2 - 6)^2 = 16
(4 - 6)^2 = 4
(6 - 6)^2 = 0
(8 - 6)^2 = 4
(10 - 6)^2 = 16
Calculate the mean of these squared differences:
(16 + 4 + 0 + 4 + 16) / 5 = 40 / 5 = 8
The variance of the dataset is 8.
Calculate the standard deviation:
sqrt(8) ≈ 2.83
The standard deviation of the dataset is approximately 2.83.

To calculate the mean deviation for a dataset, follow these steps:
Calculate the mean (average) of the dataset.
Subtract the mean from each data point and take the absolute value of the result.
Calculate the mean of these absolute differences.
Let's calculate the mean deviation for the given dataset: 3, 5, 8, 10, 12
Calculate the mean:
(3 + 5 + 8 + 10 + 12) / 5 = 38 / 5 = 7.6
Subtract the mean from each data point and take the absolute value of the result:
|3 - 7.6| = 4.6
|5 - 7.6| = 2.6
|8 - 7.6| = 0.4
|10 - 7.6| = 2.4
|12 - 7.6| = 4.4
Calculate the mean of these absolute differences:
(4.6 + 2.6 + 0.4 + 2.4 + 4.4) / 5 = 14.4 / 5 = 2.88
The mean deviation of the dataset is 2.88.

Standard deviation: $$\sqrt{25} = 5$$

Range: $$20 - 1 = 19$$

To calculate the interquartile range (IQR) of a dataset, you first need to find the values of the first and third quartiles.
The first quartile (Q1) is the value that separates the lowest 25% of the data from the highest 75% of the data.
The third quartile (Q3) is the value that separates the lowest 75% of the data from the highest 25% of the data.
To find the values of Q1 and Q3, we first need to put the data in order from lowest to highest:
1, 3, 3, 6, 8, 9
The median is the middle value of the dataset when it is ordered. In this case, the median is 5.
To find Q1, we need to find the median of the lower half of the dataset. That is, we need to find the median of 1, 3, and 3. The median of this subset is 3.
To find Q3, we need to find the median of the upper half of the dataset. That is, we need to find the median of 6, 8, and 9. The median of this subset is 8.
Now that we have Q1 and Q3, we can calculate the interquartile range (IQR) by subtracting Q1 from Q3:
IQR = Q3 - Q1 = 8 - 3 = 5
Therefore, the interquartile range (IQR) of the dataset 1, 3, 3, 6, 8, 9 is 5.

Variance: $$3^2 = 9$$

To calculate the mean deviation, we first need to find the mean of the dataset:
Mean = (12 + 15 + 18 + 21 + 24) / 5 = 18 Next, we find the deviation of each data point from the mean:
12 - 18 = -6, 15 - 18 = -3, 18 - 18 = 0, 21 - 18 = 3, 24 - 18 = 6
To calculate the mean deviation, we take the absolute value of each deviation, add them up, and divide by the number of data points:
Mean Deviation = (| -6 | + | -3 | + | 0 | + | 3 | + | 6 |) / 5 = (6 + 3 + 0 + 3 + 6) / 5 = 18 / 5 = 3.6
Therefore, the mean deviation for the given dataset is 3.6.

To calculate the range, we need to find the difference between the largest and smallest values in the dataset:
Range = Largest value - Smallest value = 15 - 5 = 10
To calculate the variance and standard deviation, we first need to find the mean (average) of the dataset:
Mean = (5 + 8 + 10 + 12 + 15) / 5 = 10
Next, we need to find the difference between each value and the mean:
5 - 10 = -5 8 - 10 = -2 10 - 10 = 0 12 - 10 = 2 15 - 10 = 5
We then square each of these differences:
(-5)^2 = 25 (-2)^2 = 4 0^2 = 0 2^2 = 4 5^2 = 25
The sum of these squared differences is:
25 + 4 + 0 + 4 + 25 = 58
To calculate the variance, we divide this sum by the number of values in the dataset minus 1:
Variance = 58 / 4 = 14.5
Finally, we can calculate the standard deviation by taking the square root of the variance:
Standard deviation = sqrt(14.5) = 3.81 (rounded to two decimal places)
Therefore, the range is 10, the variance is 14.5, and the standard deviation is 3.81 for the given dataset.

To calculate the covariance between two variables X and Y, we can use the following formula:
Cov(X,Y) = Σ [(Xi - μX) * (Yi - μY)] / (n - 1)
where
Σ represents the sum of a series
Xi is the ith value of X
Yi is the ith value of Y
μX is the mean of X
μY is the mean of Y
n is the number of observations
Using this formula, we can calculate the covariance for the given data set as follows:
Find the means of X and Y
μX = (2 + 4 + 6 + 8 + 10) / 5 = 6
μY = (5 + 8 + 11 + 14 + 17) / 5 = 11
Calculate the deviations from the mean for each observation
For X: Xi - μX = (-4, -2, 0, 2, 4)
For Y: Yi - μY = (-6, -3, 0, 3, 6)
Multiply the deviations for each observation and add them up
Σ [(Xi - μX) * (Yi - μY)] = (-4 * -6) + (-2 * -3) + (0 * 0) + (2 * 3) + (4 * 6) = 60
Divide by (n - 1) Cov(X,Y) = 60 / (5 - 1) = 15
Therefore, the covariance between X and Y for this data set is 15.

To calculate the correlation coefficient, we can use the formula:
r = cov(X, Y) / (std(X) * std(Y))
where cov(X, Y) is the covariance between X and Y, and std(X) and std(Y) are the standard deviations of X and Y, respectively.
Calculate the variance and standard deviation of X and Y:
Variance of X = Σ[(xi - mean of X)^2] / (n - 1) = [(-4)^2 + (-2)^2 + (0)^2 + (2)^2 + (4)^2] / 4 = (16 + 4 + 0 + 4 + 16) / 4 = 40 / 4 = 10
Standard deviation of X (SD_X) = sqrt(Variance of X) = sqrt(10) ≈ 3.16
Variance of Y = Σ[(yi - mean of Y)^2] / (n - 1) = [(-6)^2 + (-3)^2 + (0)^2 + (3)^2 + (6)^2] / 4 = (36 + 9 + 0 + 9 + 36) / 4 = 90 / 4 = 22.5
Standard deviation of Y (SD_Y) = sqrt(Variance of Y) = sqrt(22.5) ≈ 4.74
Divide the covariance by the product of the standard deviations of X and Y:
Correlation coefficient (r) = Cov(X,Y) / (SD_X * SD_Y) = 15 / (3.16 * 4.74) ≈ 1.00
The corrected correlation coefficient for the given bivariate data set is approximately 1.00, which indicates a perfect positive linear relationship between X and Y.

To calculate the covariance between two variables, we can use the following formula:
cov(X,Y) = [Σ(x - μx)(y - μy)] / (n - 1)
where X and Y are the two variables, μx and μy are their respective means, and n is the number of observations.
First, we need to calculate the means of X and Y:
μx = (1+2+3+4+5) / 5 = 3 μy = (10+8+6+4+2) / 5 = 6
Next, we can calculate the deviations of X and Y from their respective means:
X deviations: -2, -1, 0, 1, 2 Y deviations: 4, 2, 0, -2, -4
Multiplying these deviations together and summing them up, we get:
Σ(x - μx)(y - μy) = (-2 * 4) + (-1 * 2) + (0 * 0) + (1 * -2) + (2 * -4) = -20
Finally, we can plug in the values into the covariance formula:
cov(X,Y) = [Σ(x - μx)(y - μy)] / (n - 1) = -20 / 4 = -5
Therefore, the covariance between X and Y is -5.

To calculate the correlation coefficient (specifically, the Pearson correlation coefficient) between the variables X and Y in the previous example, follow these steps:
Calculate the covariance between X and Y (already done in the previous answer: -5).
Calculate the standard deviation of X and the standard deviation of Y.
Divide the covariance by the product of the standard deviations of X and Y.
Given the bivariate data set:
X: 1, 2, 3, 4, 5
Y: 10, 8, 6, 4, 2
We already have the covariance: Cov(X, Y) = -5.
Now let's calculate the standard deviations:
Calculate the standard deviation of X and Y:
Variance of X = Σ[(xi - mean of X)^2] / (n - 1) = [(-2)^2 + (-1)^2 + (0)^2 + (1)^2 + (2)^2] / 4 = (4 + 1 + 0 + 1 + 4) / 4 = 10 / 4 = 2.5
Standard deviation of X (SD_X) = sqrt(Variance of X) = sqrt(2.5) ≈ 1.58
Variance of Y = Σ[(yi - mean of Y)^2] / (n - 1) = [(4)^2 + (2)^2 + (0)^2 + (-2)^2 + (-4)^2] / 4 = (16 + 4 + 0 + 4 + 16) / 4 = 40 / 4 = 10
Standard deviation of Y (SD_Y) = sqrt(Variance of Y) = sqrt(10) ≈ 3.16
Divide the covariance by the product of the standard deviations of X and Y:
Correlation coefficient (r) = Cov(X, Y) / (SD_X * SD_Y) = -5 / (1.58 * 3.16) ≈ -1
The correlation coefficient for the given bivariate data set is approximately -1, which indicates a perfect negative linear relationship between X and Y.

This indicates a strong positive relationship between the two variables.

To calculate the least squares regression line, we need to find the equation of the line that minimizes the sum of the squared residuals (vertical distances between the observed points and the line).
Step 1: Calculate the mean of X and Y
x̄ = (1 + 2 + 3 + 4 + 5) / 5 = 3
ȳ = (2 + 4 + 6 + 8 + 10) / 5 = 6
Step 2: Calculate the deviations from the mean for X and Y
xi - x̄: -2, -1, 0, 1, 2
yi - ȳ: -4, -2, 0, 2, 4
Step 3: Calculate the slope of the regression line
b = Σ(xi - x̄)(yi - ȳ) / Σ(xi - x̄)^2
b = (-2*(-4) + (-1)*(-2) + 0*0 + 1*2 + 2*4) / (-2^2 + (-1)^2 + 0^2 + 1^2 + 2^2)
b = 2
Step 4: Calculate the y-intercept of the regression line
a = ȳ - b(x̄)
a = 6 - 2(3)
a = 0
Step 5: Write the equation of the regression line
ŷ = a + bx
ŷ = 0 + 2x
The least squares regression line for the given data set is ŷ = 2x.

First, find the means of X and Y:
$$\bar{x} = \frac{1 + 3 + 5 + 7 + 9}{5} = 5$$
$$\bar{y} = \frac{2 + 6 + 10 + 14 + 18}{5} = 10$$

Calculate the slope (b):
$$b = \frac{\sum_{i=1}^5 (x_i - \bar{x})(y_i - \bar{y})}{\sum_{i=1}^5 (x_i - \bar{x})^2}$$ $$= \frac{(1 - 5)(2 - 10) + (3 - 5)(6 - 10) + (5 - 5)(10 - 10)}{(1 - 5)^2 + (3 - 5)^2 + (5 - 5)^2 + (7 - 5)^2 + (9 - 5)^2}$$ $$+\frac{(7 - 5)(14 - 10) + (9 - 5)(18 - 10)}{(1 - 5)^2 + (3 - 5)^2 + (5 - 5)^2 + (7 - 5)^2 + (9 - 5)^2}$$ $$= \frac{80}{40} = 2$$

Calculate the y-intercept (a):
$$a = \bar{y} - b\bar{x} = 10 - 2(5) = 0$$

Regression line: y = 2x

To calculate the coefficient of determination (also known as R-squared), we first need to calculate the correlation coefficient (r) between X and Y.
The formula for r is:
r = [ n∑(XY) - (∑X)(∑Y) ] / [ √[n∑X^2 - (∑X)^2] √[n∑Y^2 - (∑Y)^2] ]
where n is the number of data points, ∑ is the sum of the values, X and Y represent the variables, and X^2 and Y^2 represent the squared values of X and Y, respectively.
Using the given data set, we can calculate r as follows:
n = 5
∑X = 15
∑Y = 30
∑(XY) = 70
∑X^2 = 55
∑Y^2 = 220
Substituting these values into the formula for r, we get:
r = [ (5)(70) - (15)(30) ] / [ √[(5)(55) - (15)^2] √[(5)(220) - (30)^2] ]
r = 1.00
Since the correlation coefficient is 1.00, this indicates a perfect positive correlation between X and Y. Therefore, the coefficient of determination is equal to 1.00^2 = 1.00.
Therefore, the coefficient of determination for the given data set is 1.00, indicating that all of the variation in Y can be explained by the linear relationship with X.

To calculate the correlation coefficient (specifically, the Pearson correlation coefficient) between heights and weights, follow these steps:
Calculate the mean of heights and weights.
Calculate the covariance between heights and weights.
Calculate the standard deviation of heights and the standard deviation of weights.
Divide the covariance by the product of the standard deviations of heights and weights.
Given the bivariate data set:
Heights: 62, 65, 68, 70, 72
Weights: 120, 130, 150, 160, 170
Calculate the mean of heights and weights:
Mean of Heights = (62 + 65 + 68 + 70 + 72) / 5 = 337 / 5 = 67.4
Mean of Weights = (120 + 130 + 150 + 160 + 170) / 5 = 730 / 5 = 146
Calculate the covariance between heights and weights:
First, calculate the differences between each data point and its mean:
Differences in Heights: -5.4, -2.4, 0.6, 2.6, 4.6
Differences in Weights: -26, -16, 4, 14, 24
Now, multiply the corresponding differences for each data point and sum them:
Sum of products = (-5.4 * -26) + (-2.4 * -16) + (0.6 * 4) + (2.6 * 14) + (4.6 * 24) = 140.4 + 38.4 + 2.4 + 36.4 + 110.4 = 328
Next, divide the sum of products by (n-1) for sample covariance:
Sample covariance = 328 / (5 - 1) = 328 / 4 = 82
Calculate the standard deviation of heights and weights:
Variance of Heights = Σ[(height - mean of heights)^2] / (n - 1) = [(-5.4)^2 + (-2.4)^2 + (0.6)^2 + (2.6)^2 + (4.6)^2] / 4
= (29.16 + 5.76 + 0.36 + 6.76 + 21.16) / 4 = 63.2 / 4 = 15.8 Standard deviation of Heights (SD_H) = sqrt(Variance of Heights) = sqrt(15.8) ≈ 3.98
Variance of Weights = Σ[(weight - mean of weights)^2] / (n - 1) = [(-26)^2 + (-16)^2 + (4)^2 + (14)^2 + (24)^2] / 4
= (676 + 256 + 16 + 196 + 576) / 4 = 1720 / 4 = 430 Standard deviation of Weights (SD_W) = sqrt(Variance of Weights) = sqrt(430) ≈ 20.74
Divide the covariance by the product of the standard deviations of heights and weights:
Correlation coefficient (r) = Cov(Heights, Weights) / (SD_H * SD_W) = 82 / (3.98 * 20.74) ≈ 0.987
The correlation coefficient between heights and weights is approximately 0.987, which indicates a strong positive linear relationship between the two variables.

Regression line: y = -166.89 + 4.77x
To calculate the least squares regression line, we need to find the equation of the line that best fits the data by minimizing the sum of the squared distances between the data points and the line. The equation of the line can be written as:
y = a + bx
where y is the dependent variable (in this case, the values in Y), x is the independent variable (in this case, the values in X), a is the y-intercept, and b is the slope of the line.
To find the values of a and b that minimize the sum of the squared distances, we can use the following formulas:
b = r * (Sy / Sx) a = y̅ - b * x̅
where r is the correlation coefficient, Sy is the standard deviation of Y, Sx is the standard deviation of X, y̅ is the mean of Y, and x̅ is the mean of X.
First, we need to calculate the means and standard deviations of X and Y:
x̅ = (62 + 65 + 68 + 70 + 72) / 5 = 67.4, y̅ = (120 + 130 + 150 + 160 + 170) / 5 = 146
Sx = sqrt(((62 - 67.4)^2 + (65 - 67.4)^2 + (68 - 67.4)^2 + (70 - 67.4)^2 + (72 - 67.4)^2) / 4) = 3.97
Sy = sqrt(((120 - 146)^2 + (130 - 146)^2 + (150 - 146)^2 + (160 - 146)^2 + (170 - 146)^2) / 4) = 20.74
The correlation coefficient is : 0.987
Finally, we can calculate the slope and y-intercept:
Slope: b = 0.987 * (20.74 / 3.97) = 5.16 Intercept: a = 146 - 5.16 * 67.4 = -201.78
Therefore, the equation of the least squares regression line is:
y = -201.78 + 5.16x
This means that for every one unit increase in X, we expect Y to increase by 5.16 units.

To calculate a 99% confidence interval for the population mean, you can use the formula for a confidence interval with a sample mean:
CI = x̄ ± z * (σ / √n)
Where:
CI = confidence interval
x̄ = sample mean
z = z-score (critical value) corresponding to the desired confidence level
σ = sample standard deviation
n = sample size
In this case:
x̄ = 110 liters (sample mean)
σ = 15 liters (sample standard deviation)
n = 225 (sample size)
To find the z-score for a 99% confidence interval, you can use a z-table or an online calculator. The z-score for a 99% confidence interval is approximately 2.576.
Now, plug the values into the formula:
CI = 110 ± 2.576 * (15 / √225)
Calculate the standard error (σ / √n):
Standard error = 15 / √225 ≈ 1
Now, calculate the margin of error (z * standard error):
Margin of error = 2.576 * 1 ≈ 2.576
Finally, calculate the confidence interval:
Lower limit = 110 - 2.576 ≈ 107.424
Upper limit = 110 + 2.576 ≈ 112.576
So, the 99% confidence interval for the population mean is approximately (107.424, 112.576) liters. This means that you can be 99% confident that the true population mean of monthly fuel consumption lies within this range.

To calculate a 90% confidence interval for the population proportion, you can use the formula for a confidence interval with a sample proportion:
CI = p̂ ± z * √(p̂ * (1 - p̂) / n)
Where:
CI = confidence interval
p̂ = sample proportion (successes / sample size)
z = z-score (critical value) corresponding to the desired confidence level
n = sample size
In this case:
p̂ = 175 / 500 = 0.35 (sample proportion)
n = 500 (sample size)
To find the z-score for a 90% confidence interval, you can use a z-table or an online calculator. The z-score for a 90% confidence interval is approximately 1.645.
Now, plug the values into the formula:
CI = 0.35 ± 1.645 * √(0.35 * (1 - 0.35) / 500)
Calculate the standard error (SE = √(p̂ * (1 - p̂) / n)):
Standard error = √(0.35 * (1 - 0.35) / 500) ≈ 0.0213
Now, calculate the margin of error (z * standard error):
Margin of error = 1.645 * 0.0213 ≈ 0.0350
Finally, calculate the confidence interval:
Lower limit = 0.35 - 0.0350 ≈ 0.3150
Upper limit = 0.35 + 0.0350 ≈ 0.3850
So, the 90% confidence interval for the population proportion is approximately (0.3150, 0.3850). This means that you can be 90% confident that the true population proportion of people who prefer brand A over brand B lies within this range.

Let's compute the confidence interval using the z-distribution:
CI = x̄ ± z * (σ / √n)
Where:
CI = confidence interval
x̄ = sample mean
z = z-score (critical value) corresponding to the desired confidence level
σ = sample standard deviation
n = sample size
In this case:
x̄ = 1200 hours (sample mean)
σ = 200 hours (sample standard deviation)
n = 36 (sample size)
To find the z-score for a 95% confidence interval, you can use a z-table or an online calculator. The z-score for a 95% confidence interval is approximately 1.96.
Now, plug the values into the formula:
CI = 1200 ± 1.96 * (200 / √36)
Calculate the standard error (σ / √n):
Standard error = 200 / √36 ≈ 33.33
Now, calculate the margin of error (z * standard error):
Margin of error = 1.96 * 33.33 ≈ 65.33
Finally, calculate the confidence interval:
Lower limit = 1200 - 65.33 ≈ 1134.67
Upper limit = 1200 + 65.33 ≈ 1265.33
So, the 95% confidence interval for the population mean is approximately (1134.67, 1265.33) hours. This means that you can be 95% confident that the true population mean of light bulb lifetimes lies within this range.

Answer:
99% confidence interval: $$(150 - 2.576 \frac{20}{\sqrt{49}}, 150 + 2.576 \frac{20}{\sqrt{49}}) = (142.64, 157.36)$$
Details and explanations
To calculate a 99% confidence interval for the population mean, you can use the formula for a confidence interval with a sample mean:
CI = x̄ ± z * (σ / √n)
Where:
CI = confidence interval
x̄ = sample mean
z = z-score (critical value) corresponding to the desired confidence level
σ = sample standard deviation
n = sample size
In this case:
x̄ = 150 grams (sample mean)
σ = 20 grams (sample standard deviation)
n = 49 (sample size)
To find the z-score for a 99% confidence interval, you can use a z-table or an online calculator. The z-score for a 99% confidence interval is approximately 2.576.
Now, plug the values into the formula:
CI = 150 ± 2.576 * (20 / √49)
Calculate the standard error (σ / √n):
Standard error = 20 / √49 ≈ 2.857
Now, calculate the margin of error (z * standard error):
Margin of error = 2.576 * 2.857 ≈ 7.357
Finally, calculate the confidence interval:
Lower limit = 150 - 7.357 ≈ 142.643
Upper limit = 150 + 7.357 ≈ 157.357
So, the 99% confidence interval for the population mean is approximately (142.643, 157.357) grams. This means that you can be 99% confident that the true population mean of apple weights lies within this range.

Answer:
90% confidence interval: $$(0.65 - 1.645 \sqrt{\frac{0.65(1 - 0.65)}{800}}, 0.65 + 1.645 \sqrt{\frac{0.65(1 - 0.65)}{800}}) = (0.62, 0.678)$$
Details and explanations:
To calculate a 90% confidence interval for the population proportion, you can use the formula for a confidence interval with a sample proportion:
CI = p̂ ± z * √(p̂ * (1 - p̂) / n)
Where:
CI = confidence interval
p̂ = sample proportion (successes / sample size)
z = z-score (critical value) corresponding to the desired confidence level
n = sample size
In this case:
p̂ = 520 / 800 = 0.65 (sample proportion)
n = 800 (sample size)
To find the z-score for a 90% confidence interval, you can use a z-table or an online calculator. The z-score for a 90% confidence interval is approximately 1.645.
Now, plug the values into the formula:
CI = 0.65 ± 1.645 * √(0.65 * (1 - 0.65) / 800)
Calculate the standard error (SE = √(p̂ * (1 - p̂) / n)):
Standard error = √(0.65 * (1 - 0.65) / 800) ≈ 0.0169
Now, calculate the margin of error (z * standard error):
Margin of error = 1.645 * 0.0164 ≈ 0.0278
Finally, calculate the confidence interval:
Lower limit = 0.65 - 0.0278 ≈ 0.6222
Upper limit = 0.65 + 0.0278 ≈ 0.6778
So, the 90% confidence interval for the population proportion is approximately (0.6230, 0.6770). This means that you can be 90% confident that the true population proportion of people who prefer coffee over tea lies within this range.

Answer:
95% confidence interval: $$(7.5 - 1.96 \frac{1.5}{\sqrt{81}}, 7.5 + 1.96 \frac{1.5}{\sqrt{81}}) = (7.17, 7.83)$$
Details and explanations:
To calculate a 95% confidence interval for the population mean, you can use the formula for a confidence interval with a sample mean:
CI = x̄ ± z * (σ / √n)
Where:
CI = confidence interval
x̄ = sample mean
z = z-score (critical value) corresponding to the desired confidence level
σ = sample standard deviation
n = sample size
In this case:
x̄ = 7.5 cm (sample mean)
σ = 1.5 cm (sample standard deviation)
n = 81 (sample size)
To find the z-score for a 95% confidence interval, you can use a z-table or an online calculator. The z-score for a 95% confidence interval is approximately 1.96.
Now, plug the values into the formula:
CI = 7.5 ± 1.96 * (1.5 / √81)
Calculate the standard error (σ / √n):
Standard error = 1.5 / √81 ≈ 0.1667
Now, calculate the margin of error (z * standard error):
Margin of error = 1.96 * 0.1667 ≈ 0.3271
Finally, calculate the confidence interval:
Lower limit = 7.5 - 0.3271 ≈ 7.1729
Upper limit = 7.5 + 0.3271 ≈ 7.8271
So, the 95% confidence interval for the population mean is approximately (7.1729, 7.8271) cm. This means that you can be 95% confident that the true population mean of orange diameters lies within this range.

Answer:
99% confidence interval: $$(16 - 2.576 \frac{4}{\sqrt{121}}, 16 + 2.576 \frac{4}{\sqrt{121}}) = (15.06, 16.94)$$
Details and explanations:
To calculate a 99% confidence interval for the population mean, you can use the formula for a confidence interval with a sample mean:
CI = x̄ ± z * (σ / √n)
Where:
CI = confidence interval
x̄ = sample mean
z = z-score (critical value) corresponding to the desired confidence level
σ = sample standard deviation
n = sample size
In this case:
x̄ = 16 hours (sample mean)
σ = 4 hours (sample standard deviation)
n = 121 (sample size)
To find the z-score for a 99% confidence interval, you can use a z-table or an online calculator. The z-score for a 99% confidence interval is approximately 2.576.
Now, plug the values into the formula:
CI = 16 ± 2.576 * (4 / √121)
Calculate the standard error (σ / √n):
Standard error = 4 / √121 ≈ 0.364
Now, calculate the margin of error (z * standard error):
Margin of error = 2.576 * 0.364 ≈ 0.938
Finally, calculate the confidence interval:
Lower limit = 16 - 0.938 ≈ 15.062
Upper limit = 16 + 0.938 ≈ 16.938
So, the 99% confidence interval for the population mean is approximately (15.062, 16.938) hours. This means that you can be 99% confident that the true population mean of smartphone battery life lies within this range.

Answer:
95% confidence interval: $$(0.25 - 1.96 \sqrt{\frac{0.25(1 - 0.25)}{1000}}, 0.25 + 1.96 \sqrt{\frac{0.25(1 - 0.25)}{1000}}) = (0.223, 0.277)$$
Details and explanations:
To calculate a 95% confidence interval for the population proportion, you can use the formula for a confidence interval with a sample proportion:
CI = p̂ ± z * √(p̂ * (1 - p̂) / n)
Where:
CI = confidence interval
p̂ = sample proportion (successes / sample size)
z = z-score (critical value) corresponding to the desired confidence level
n = sample size
In this case:
p̂ = 250 / 1000 = 0.25 (sample proportion)
n = 1000 (sample size)
To find the z-score for a 95% confidence interval, you can use a z-table or an online calculator. The z-score for a 95% confidence interval is approximately 1.96.
Now, plug the values into the formula:
CI = 0.25 ± 1.96 * √(0.25 * (1 - 0.25) / 1000)
Apologies for the error in my previous calculations. Let's correct that and find the confidence interval again.
Calculate the standard error (SE = √(p̂ * (1 - p̂) / n)):
Standard error = √(0.25 * (1 - 0.25) / 1000) ≈ 0.01369
Now, calculate the margin of error (z * standard error):
Margin of error = 1.96 * 0.01369 ≈ 0.02683
Finally, calculate the confidence interval:
Lower limit = 0.25 - 0.02683 ≈ 0.22317
Upper limit = 0.25 + 0.02683 ≈ 0.27683
So, the 95% confidence interval for the population proportion is approximately (0.22317, 0.27683). This means that you can be 95% confident that the true population proportion of people who prefer online shopping over in-store shopping lies within this range.

Answer:
95% confidence interval: $$(950 - 1.96 \frac{150}{\sqrt{64}}, 950 + 1.96 \frac{150}{\sqrt{64}}) = (913.25, 986.75)$$
Details and explanations:
To calculate a 95% confidence interval for the population mean, you can use the formula for a confidence interval with a sample mean:
CI = x̄ ± z * (σ / √n)
Where:
CI = confidence interval
x̄ = sample mean
z = z-score (critical value) corresponding to the desired confidence level
σ = sample standard deviation
n = sample size
In this case:
x̄ = $950 (sample mean)
σ = $150 (sample standard deviation)
n = 64 (sample size)
To find the z-score for a 95% confidence interval, you can use a z-table or an online calculator. The z-score for a 95% confidence interval is approximately 1.96.
Now, plug the values into the formula:
CI = 950 ± 1.96 * (150 / √64)
Calculate the standard error (σ / √n):
Standard error = 150 / √64 ≈ 18.75
Now, calculate the margin of error (z * standard error): Margin of error = 1.96 * 18.75 ≈ 36.75
Finally, calculate the confidence interval:
Lower limit = 950 - 36.75 ≈ 913.25
Upper limit = 950 + 36.75 ≈ 986.75
So, the 95% confidence interval for the population mean is approximately ($913.25, $986.75). This means that you can be 95% confident that the true population mean of laptop prices lies within this range.

Null hypothesis: $$H_0: \mu \leq 2$$
Alternative hypothesis: $$H_1: \mu > 2$$
Test statistic: $$z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} = \frac{2.5 - 2}{\frac{0.8}{\sqrt{36}}} = 3.75$$
Critical value: $$z_{0.05} = 1.645$$
Since $$z > z_{0.05}$$, we reject the null hypothesis. There is sufficient evidence to support the claim that high school students spend more than 2 hours on homework per day.

Null hypothesis: $$H_0: \mu = 300$$
Alternative hypothesis: $$H_1: \mu \neq 300$$
Test statistic: $$t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{290 - 300}{\frac{20}{\sqrt{25}}} = -2.5$$
Critical values: $$t_{0.005, 24} = \pm 2.797$$
Since $$-2.797 < t < 2.797$$, we fail to reject the null hypothesis. There is not sufficient evidence to reject the claim that the electric car has a mean range of 300 miles on a full charge.

Null hypothesis: $$H_0: \mu = 68$$
Alternative hypothesis: $$H_1: \mu \neq 68$$
Test statistic: $$z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} = \frac{67 - 68}{\frac{3}{\sqrt{100}}} = -3.33$$
Critical values: $$z_{0.025} = \pm 1.96$$
Since $$z < -1.96$$ or $$z > 1.96$$, we reject the null hypothesis. There is sufficient evidence to support the claim that the average height of male students in the high school is different from 5'8".

Null hypothesis: $$H_0: \mu \geq 3.5$$
Alternative hypothesis: $$H_1: \mu < 3.5$$
Test statistic: $$t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{3.2 - 3.5}{\frac{0.4}{\sqrt{16}}} = -3$$
Critical value: $$t_{0.1, 15} = -1.341$$
Since $$t < t_{0.1, 15}$$, we reject the null hypothesis. There is sufficient evidence to reject the claim that the mean weight of their chocolate bars is at least 3.5 ounces.

To test this claim, we can follow the steps of hypothesis testing:
State the null and alternative hypotheses:
Null Hypothesis (H_0): The proportion of students who pass the math exam is 0.8. (p = 0.8)
Alternative Hypothesis (H_a): The proportion of students who pass the math exam is not 0.8. (p ≠ 0.8)
Choose the appropriate test statistic:
Since we have a large sample size (n = 200) and we are testing proportions, we can use a Z-test.
Calculate the test statistic:
First, find the sample proportion (p̂): p̂ = number of successes / sample size = 150 / 200 = 0.75
$$Z = (p̂ - p_0) / sqrt((p_0 * (1 - p_0)) / n) = (0.75 - 0.8) / sqrt((0.8 * 0.2) / 200)$$ $$ = (-0.05) / 0.0283 ≈ -1.77$$
Determine the P-value:
Since this is a two-tailed test (p ≠ 0.8), we need to find the P-value by looking up the Z-score in a standard normal table or using a calculator. For a Z-score of -1.77, the P-value is approximately 0.0764.
Compare the P-value to the significance level:
Given the significance level α = 0.05, since the P-value (0.0764) > α, we fail to reject the null hypothesis.
Conclusion: There is not sufficient evidence to reject the school district's claim that the proportion of students who pass the standardized math exam is 0.8 at a 0.05 significance level.

To test this claim, we can follow the steps of hypothesis testing:
State the null and alternative hypotheses:
Null Hypothesis (H_0): The proportion of customers who are satisfied is at least 90%. (p ≥ 0.9)
Alternative Hypothesis (H_a): The proportion of customers who are satisfied is less than 90%. (p < 0.9)
Choose the appropriate test statistic:
Since we have a large sample size (n = 100) and we are testing proportions, we can use a Z-test.
Calculate the test statistic:
First, find the sample proportion (p̂): p̂ = number of satisfied customers / sample size = 85 / 100 = 0.85
$$Z = (p̂ - p_0) / sqrt((p_0 * (1 - p_0)) / n) = (0.85 - 0.9) / sqrt((0.9 * 0.1) / 100)$$ $$ = (-0.05) / 0.03 ≈ -1.67$$ Determine the P-value:
Since this is a one-tailed test (p < 0.9), we need to find the P-value by looking up the Z-score in a standard normal table or using a calculator. For a Z-score of -1.67, the P-value is approximately 0.0475.
Compare the P-value to the significance level:
Given the significance level α = 0.1, since the P-value (0.0475) < α, we reject the null hypothesis.
Conclusion: There is sufficient evidence to reject the company's claim that at least 90% of their customers are satisfied with their products at a 0.1 significance level.

Null hypothesis: $$H_0: \mu = 10,000$$
Alternative hypothesis: $$H_1: \mu \neq 10,000$$
Test statistic: $$t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{9,600 - 10,000}{\frac{800}{\sqrt{9}}} = -1.5$$
Critical values: $$t_{0.025, 8} = \pm 2.306$$
Since $$-2.306 < t < 2.306$$, we fail to reject the null hypothesis. There is not sufficient evidence to reject the claim that the new lightbulbs last an average of 10,000 hours.

Null hypothesis: $$H_0: \mu \geq 15$$
Alternative hypothesis: $$H_1: \mu < 15$$
Test statistic: $$t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{13 - 15}{\frac{4}{\sqrt{20}}} = -2.236$$
Critical value: $$t_{0.1, 19} = -1.328$$
Since $$t < t_{0.1, 19}$$, we reject the null hypothesis. There is sufficient evidence to support the claim that the average waiting time for a table is less than 15 minutes.

Null hypothesis: $$H_0: \mu \geq 50$$
Alternative hypothesis: $$H_1: \mu < 50$$
Test statistic: $$t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{48 - 50}{\frac{6}{\sqrt{15}}} = -1.443$$
Critical value: $$t_{0.05, 14} = -1.761$$
Since $$t > t_{0.05, 14}$$, we fail to reject the null hypothesis. There is not sufficient evidence to reject the claim that the mean lifetime of their batteries is at least 50 hours.

Null hypothesis: $$H_0: p \geq 0.1$$
Alternative hypothesis: $$H_1: p < 0.1$$
Test statistic: $$z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} = \frac{0.125 - 0.1}{\sqrt{\frac{0.1(1 - 0.1)}{200}}} = 1.667$$
Critical value: $$z_{0.05} = -1.645$$
Since $$z > z_{0.05}$$, we fail to reject the null hypothesis. There is not sufficient evidence to support the claim that less than 10% of their students drop out before graduating.

Null hypothesis: $$H_0: \mu \leq 2$$
Alternative hypothesis: $$H_1: \mu > 2$$
Test statistic: $$z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} = \frac{2.5 - 2}{\frac{0.8}{\sqrt{36}}} = 3.375$$
Critical value: $$z_{0.05} = 1.645$$
Since $$z > z_{0.05}$$, we reject the null hypothesis. There is sufficient evidence to support the claim that high school students spend more than 2 hours on homework per day.

Null hypothesis: $$H_0: \mu = 300$$
Alternative hypothesis: $$H_1: \mu \neq 300$$
Test statistic: $$t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{290 - 300}{\frac{20}{\sqrt{25}}} = -2.5$$
Critical values: $$t_{0.005, 24} = \pm 2.797$$
Since $$-2.797 < t < 2.797$$, we fail to reject the null hypothesis. There is not sufficient evidence to reject the claim that the electric car has a mean range of 300 miles on a full charge.

Null hypothesis: $$H_0: \mu = 68$$
Alternative hypothesis: $$H_1: \mu \neq 68$$
Test statistic: $$z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} = \frac{67 - 68}{\frac{3}{\sqrt{100}}} = -3.33$$
Critical values: $$z_{0.025} = \pm 1.96$$
Since $$z < -1.96$$ or $$z > 1.96$$, we reject the null hypothesis. There is sufficient evidence to support the claim that the average height of male students in the high school is different from 5'8".

Null hypothesis: $$H_0: \mu \geq 3.5$$
Alternative hypothesis: $$H_1: \mu < 3.5$$
Test statistic: $$t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{3.2 - 3.5}{\frac{0.4}{\sqrt{16}}} = -3$$
Critical value: $$t_{0.1, 15} = -1.341$$
Since $$t < t_{0.1, 15}$$, we reject the null hypothesis. There is sufficient evidence to reject the claim that the mean weight of their chocolate bars is at least 3.5 ounces.

Null hypothesis: $$H_0: p = 0.8$$
Alternative hypothesis: $$H_1: p \neq 0.8$$
Test statistic: $$z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} = \frac{0.75 - 0.8}{\sqrt{\frac{0.8(1 - 0.8)}{200}}} = -1.667$$
Critical values: $$z_{0.025} = \pm 1.96$$
Since $$-1.96 < z < 1.96$$, we fail to reject the null hypothesis. There is not sufficient evidence to reject the claim that the proportion of students who pass the standardized math exam is 0.8.

Null hypothesis: $$H_0: p \geq 0.9$$
Alternative hypothesis: $$H_1: p < 0.9$$
Test statistic: $$z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} = \frac{0.85 - 0.9}{\sqrt{\frac{0.9(1 - 0.9)}{100}}} = -3.333$$
Critical value: $$z_{0.1} = -1.282$$
Since $$z < z_{0.1}$$, we reject the null hypothesis. There is sufficient evidence to reject the claim that at least 90% of their customers are satisfied with their products.

Null hypothesis: $$H_0: \mu = 10,000$$
Alternative hypothesis: $$H_1: \mu \neq 10,000$$
Test statistic: $$t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{9,600 - 10,000}{\frac{800}{\sqrt{9}}} = -1.5$$
Critical values: $$t_{0.025, 8} = \pm 2.306$$
Since $$-2.306 < t < 2.306$$, we fail to reject the null hypothesis. There is not sufficient evidence to reject the claim that the new lightbulbs last an average of 10,000 hours.

Null hypothesis: $$H_0: \mu \geq 15$$
Alternative hypothesis: $$H_1: \mu < 15$$
Test statistic: $$t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{13 - 15}{\frac{4}{\sqrt{20}}} = -2.236$$
Critical value: $$t_{0.1, 19} = -1.328$$
Since $$t < t_{0.1, 19}$$, we reject the null hypothesis. There is sufficient evidence to support the claim that the average waiting time for a table is less than 15 minutes.

Null hypothesis: $$H_0: \mu \geq 50$$
Alternative hypothesis: $$H_1: \mu < 50$$
Test statistic: $$t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{48 - 50}{\frac{6}{\sqrt{15}}} = -1.443$$
Critical value: $$t_{0.05, 14} = -1.761$$
Since $$t > t_{0.05, 14}$$, we fail to reject the null hypothesis. There is not sufficient evidence to reject the claim that the mean lifetime of their batteries is at least 50 hours.

Null hypothesis: $$H_0: p \geq 0.1$$
Alternative hypothesis: $$H_1: p < 0.1$$
Test statistic: $$z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} = \frac{0.125 - 0.1}{\sqrt{\frac{0.1(1 - 0.1)}{200}}} = 1.667$$
Critical value: $$z_{0.05} = -1.645$$
Since $$z > z_{0.05}$$, we fail to reject the null hypothesis. There is not sufficient evidence to support the claim that less than 10% of their students drop out before graduating.