Question

Answer following questions using information from the following tables: Standard Normal Distribution Probability0.900950.99 Threshold 164 1.96 258

Answer 1

2.1) Since we know the variance of the population, we use Z-score for the confidence interval. The confidence interval is

$(\bar{X}-Z_{0.95}\frac{\sigma}{\sqrt{n}},\bar{X}+Z_{0.95}\frac{\sigma}{\sqrt{n}}) = (4.5-1.96(1.1/\sqrt{25}),4.5+1.96(1.1/\sqrt{25}))=\\ (4.0688, 4.9312)$

2.2) Since we don't know the variance of the population, we use t-statistic (with 16-1=15 degrees of freedom) for the confidence interval. The confidence interval is

$(\bar{X}-t_{0.975,15}\frac{s}{\sqrt{n}},\bar{X}+t_{0.975,15}\frac{s}{\sqrt{n}}) = (7.8-2.131(2.5/\sqrt{16}),7.8+2.131(2.5/\sqrt{16}))=\\ (6.47, 9.13)$

2.3) Given the hypothesis, we have to do a two-tailed t-test to check for the significance.

the t-statistics is

$t = \frac{\bar{X}-13.12}{(s/\sqrt{n})} = \frac{12.64-13.12}{1/\sqrt{16}} = -1.92.\\ The\:p-value\:is(for\:15\:degrees\:of\:freedom)\: p = 0.074.$

The p-value is 0.074 and is greater than 0.05 but less than 0.1. Thus, if our significance level is 90% we reject our null (and say that there is statistically significant impact. Also, if our significance level is 95%, we fail to reject our null (and cannot talk about significance of the impact)