Question

A random sample of 25 statistics examinations was taken. the average score in the sample was 76 with a standard deviation of 12. assuming the scores are normally distributed, the 99% confidence interval for the population average examination score is:

Answer 1

Since the sample size is less than 30, therefore we use the t statistic. Let us define the given variables: N = sample size = 25 X = average score = 76 s = standard deviation = 12 99% Confidence interval Degrees of freedom = n – 1 = 24 The formula for confidence interval is given as: CI = X ± t * s / sqrt N using the standard distribution table, the t value for DF = 24 and 99% CI is: t = 2.492 Therefore calculating the CI using the known values: CI = 76 ± 2.492 * 12 / sqrt 25 CI = 76 ± 5.98 CI = 70.02, 81.98 Answer: The average score ranges from 70 to 82.

Explanation:

Answer 2

Since the sample size is less than 30, therefore we use the t statistic.

Let us define the given variables:

N = sample size = 25

X = average score = 76

s = standard deviation = 12

99% Confidence interval

Degrees of freedom = n – 1 = 24

The formula for confidence interval is given as:

CI = X ± t * s / sqrt N

using the standard distribution table, the t value for DF = 24 and 99% CI is:

t = 2.492

Therefore calculating the CI using the known values:

CI = 76 ± 2.492 * 12 / sqrt 25

CI = 76 ± 5.98

CI = 70.02, 81.98

Answer: The average score ranges from 70 to 82.