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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:

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barnuts

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.

meganeads

Answer:

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.

Step-by-step explanation:

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