Question

A random 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

Answer 1

Answer:

The confidence interval is from 69.82 o 82.18

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