Arandom sample of 25 statistics examinations was selected. 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. 69.29 to 82.71 d. 70.06 to 81.94



The confidence interval is from 69.82 o 82.18

Step-by-step explanation:

Using this formula X ± Z (s/√n)

Where

X = 76 Mean

S = Standard Deviation

If Variance = 144

S = √144

S = 12

n = 25 Number of observation

Z = 2.576 The chosen Z-value from the confidence table below

Confidence Interval || Z

80%. || 1.282

85% || 1.440

90%. || 1.645

95%. || 1.960

99%. || 2.576

99.5%. || 2.807

99.9%. || 3.291

Substituting these values in the formula

Confidence Interval (CI) = 76 ± 2.576 (12/√25)

CI = 76 ± 2.576(12/5)

CI = 76 ± 2.576(2.4)

CI = 76 ± 6.1824

CI = 76 + 6.1824 ~ 76 - 6.1824

CI = 82.1824 ~ 69.8176

CI = 82.18 ~ 69.82

In other words the confidence interval is from 69.8176 to 82.1824

answer: 48

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