Question

In a class of 100 students, the average score in the final exam was 72 out...

In a class of 100 students, the average score in the final exam was 72 out of 100 with a standard deviation of 5. A statistician computes the following approximate 95% confidence interval 72 ± 1.96 · 5/ √ 100 = 72 ± 0.98. Is he correct or not? Explain.

Math   Statistics-And-Probability   
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Answer #1

Solution:

Given that,

The equation is not correct.

When standard deviation is unknown , then we use t - distribution.

From the given problem , there is a z - distribution, then the given equation is not correct.

The correct equation is,

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,99 = 1.984

72 ± 1.984 · 5/ √ 100 = 72 ± 0.992.

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