1. In order to estimate the **proportion **of office workers who listen to streamed music on a work computer on a regular basis, a sample of 1200 office workers who work at a computer was taken. Of them, 543 listen to steamed music on the computer at work. A 99% confidence interval for the **proportion **of all office workers who listen to streamed music at work is about:

(0.453-0.037, 0.453+0.037)

(0.453-0.044,0.453+0.044)

(0.453-0.028, 0.453+0.028)

(0.453-0.033, 453+0.033)

(0.453-0.001, 0.453+0.001)

The **mean **monthly salary of a **random sample **of 20 college graduates under the age of 30 was found to be $3120 with a **standard deviation **of $67. Assume that the distribution of salaries for all college graduates under the age of 30 is **normally distributed.**

(a) Construct a 90% confidence interval for the **population mean **of monthly salaries of all college graduates under the age of 30.

(2741, 3498)

(3094,3146)

(1071, 1569)

(1058, 1582)

(108, 4831)

(b) All other information remaining unchanged, which of the following will produce a **wider **interval than the 90% confidence interval constructed above?

A sample with a standard deviation of 52 instead of 67

An 80% confidence interval rather than a 90% confidence interval

A sample of size 28 instead of 20

A sample of size 24 instead of 20

A sample with a standard deviation of 72 instead of 67

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