Question

1. Now assume that instead of a sample of 365 kindergarten students, your sample consisted of 16 randomly selected kids. Find the 95% confidence interval of the mean for all kindergarteners in the US.

A study of 365 randomly-selected kindergarten students across the US revealed that they had seen an average of 3000 hours of television. If the standard deviation is 300, us distribution find the 95% confidence interval of the mean for all kindergarteners in the U 5) e the Z- 6) Now assume that instead of a sample of 365 kindergarten students, your sample consisted of 16 randomly selected kids. Find the 95% confidence interval of the mean for all kindergarteners in the US.

Answer 1

5. z value for 95% CI is 1.96 as P(-1.96<z<1.96)=0.95

So Margin of Error is $E=z*\frac{\sigma }{\sqrt{n}}=1.96*\frac{300}{\sqrt{365}} =30.777$

Hence CI is $CI=\overline{x} \pm E= 3000 \pm 30.777=(2969.223, 3030.777)$

6. t table value for 15 df is 2.131

So $E=t*\frac{\sigma }{\sqrt{n}} = 2.131*\frac{300}{\sqrt{16}} =159.825$

Hence CI is $CI=\overline{x} \pm E= 3000 \pm 159.825=(2840.175,3159.825)$