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.
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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