Question

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

Answer 1

Ans:

1)

sample proportion=543/1200=0.453

99% confidence interval for true p

=0.453+/-2.58*sqrt(0.453*(1-0.453)/1200)

=0.453+/-0.037

2)

a)df=20-1=19

critical t value=tinv(0.1,19)=1.729

90% confodenec interval for true mean

=3120+/-1.729*(67/sqrt(20))

=3120+/-26

=(3094,3146)

b)Correct option is:

A sample with a standard deviation of 72 instead of 67

Larger the standard deviation,larger the margin of error and wider the confidence interval.

(larger sample size will produce a narrower interval and lower confidence level will produce narrower interval too)