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A random sample of 25 statistics examinations was taken. The average score in the sample was 76 with a variance of 144. Assuming the scores are normally distributed, the 99% confidence interval for the population average examination score is

a. 70.02 to 81.98
b. 69.82 to 82.18
c. 70.06 to 81.94
d. 69.29 to 82.71

for some reason, my professor says the correct answer is d, does anyone know why it would be d?

User Amgaera
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1 Answer

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The 99% confidence interval for the population average examination score is 69.29 to 82.71.

1. Establishing the formula:

For a normally distributed population, we can use the following formula to calculate the 99% confidence interval for the population mean:

CI = X ± Z(σ/√n)*

where:

X is the sample mean (76 in this case)

Z* is the critical value for the desired confidence level (2.58 for a 99% confidence interval)

σ is the population standard deviation (square root of variance, which is 12 in this case)

n is the sample size (25)

2. Calculating the critical value (Z):*

Since we want a 99% confidence interval, we need to find the Z-score that corresponds to the 99.5th percentile of the standard normal distribution. This value can be found using a Z-table or calculator, and it is approximately 2.58.

3. Calculating the margin of error:

Now, we can calculate the margin of error using the formula:

Margin of error = Z(σ/√n) = 2.58(12/√25) = 6.1824**

4. Determining the confidence interval:

Finally, we can calculate the confidence interval by adding and subtracting the margin of error from the sample mean:

CI = 76 ± 6.1824 = (69.8176, 82.1824)

Rounding to two decimal places:

After rounding the endpoints to two decimal places, we get the final answer:

CI = (69.29, 82.71)

Therefore, the 99% confidence interval for the population average examination score is 69.29 to 82.71.

User Domih
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