A student believes that the average grade on the statistics final examination is 87. A sample of 36 final examinations is taken. The average grade in the sample is 83.96. The sample variance is 144. Compute a 92% confidence interval: (Given t0.08 = -1.43, t0.04 = -1.80 both with 35 degree of freedom; z0.04 = -1.75, z0.08 = -1.41)
We have given that,
Sample mean =83.96
Sample standard deviation =12
Sample size =36
Level of significance=1-0.92=0.08
Degree of freedom =35
t critical value is (by using t table)=1.803
Therefore, 92% confidence interval is =( 80.354, 87.566) |
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