Question

The next three questions (Q28, Q29, and Q30) are based on the following paragraph. 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 0 to 100 point scale) on safety procedures. For that sample of employees, the mean test score was 75 points, with a sample 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).

28. To construct the 90% confidence interval, we should:

A. use t-value 1.669 from the t table, because we have the sample standard deviation.

B. use t-value 1.998 from the t table, because we have the sample standard deviation.

C. use z-value 1.645 from the z table, because we have the population standard deviation.

D. use z-value 1.960 from the z table, because we have the population standard deviation.

29. The 90% confidence interval is:

A. [71.8706, 78.1294]

B. [71.0687, 78.9412]

C. [71.2537, 78.7463]

D. [71.3725, 78.6374]

- 30. We can interpret the interval in the following manner:

A. We are 90% confident that the population mean test score of all employees is covered by the interval.

B. We are 10% confident that the population mean test score of all employees is covered by the interval.

C. We are 90% confident that the population mean test score of all employees is NOT covered by the interval.

D. The length of the interval is 90% of one.

Answer 1

Answer)

As the population s.d is not mentioned here and we are given with the sample s.d as the best estimate we will use t distribution to estimate the interval

Degrees of freedom is = n-1 = 63

For 63 dof and 90% confidence level critical value t from.t distribution is = 1.67

Margin of error (MOE) = t*s.d/√n = 1.67*15/√64 = 3.13

Interval is given by

(Mean - MOE, Mean + MOE)

[71.87, 78.13].

You can be 90% confident that the population mean (μ) falls between 71.87 and 78.13.

28)

A. use t-value 1.669 from the t table, because we have the sample standard deviation

29)

A. [71.8706, 78.1294]

30)

A. We are 90% confident that the population mean test score of all employees is covered by the interval