Question

Use the given degree of confidence and sample data to construct a confidence interval for the population mean u. Assume that the population has a normal distribution. A sociologist develops a test to measure attitudes towards public transportation, and 27 randomly selected subjects are given the test. Their mean score is 76.2 and their standard deviation is 21.4. Construct the 95% confidence interval for the mean score of all such subjects. 074.6 < u < 77.8 0 69.2 < < 83.2 064.2<u < 88.2 067.7 < p < 84.7

Answer 1

Solution :

Given that,

$\bar x$ = 76.2

s =21.4

n = Degrees of freedom = df = n - 1 = 27- 1 = 26

At 95% confidence level the t is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

  $\alpha$/ 2= 0.05 / 2 = 0.025

t$\alpha$ /2,df = t0.025,26 = 2.056 ( using student t table)

Margin of error = E = t$\alpha$/2,df * (s /$\sqrt$n)

= 2.056 * (21.4 / $\sqrt$ 27)

= 8.5

The 95% confidence interval is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

76.2 - 8.5 < $\mu$ < 76.2+ 8.5

67.7< $\mu$ < 84.7