Question

A large hospital wholesaler, as part of an assessment of workplace safety, gave a random sample of 64 of its warehouse employees a test (measured on a O to 100 point scale) on safety procedures. For that sample of employees, the mean test score was 75 points, with a standard deviation of 15 points. Determine and interpret a 90% confidence interval for the mean test score of all the company's warehouse employees. (Please keep at least four decimal places). Question 17 (1 point) The appropriate interpretation for the above mentioned confidence interval is We are 90% confident that the mean test score of the sample of employees is between 71.8706 and 78.1294 points. We are 90% confident that the population mean test score of all employees is between 71.8706 and 78.1294 points. We are 95% confident that the population mean test score of all employees is between 71.3250 and 78.6750 points. We are 90% confident that the population mean test score of all employees is between 71.9156 and 78.0844 points

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 75

sample standard deviation = s = 15

sample size = n = 64

Degrees of freedom = df = n - 1 = 64 - 1 = 63

At 90% confidence level the t is ,

$\alpha$ = 1 - 90% = 1 - 0.90 = 0.10

$\alpha$ / 2 = 0.10 / 2 = 0.05

t _{$\alpha$ /2,df} = t_0.05,63 = 1.669

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

= 1.669 * (15 / $\sqrt$ 64)

= 3.1294

The 95% confidence interval estimate of the population mean is,

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

75-3.1294 < $\mu$ < 75-3.1294

71.8706 < $\mu$ < 78.1294

(71.8706,78.1294)

We are 90% confident that the population mean test score of the sample of all employees is between 71.8706 and 78.1294 points