Question

A study reported that finger rings increase the growth of bacteria on health-care workers’ hands. Research suggests that 31 percent of health-care workers who wear rings have bacteria on one or both hands, and 27 percent of health-care workers without rings have bacteria on one or both hands. Suppose that independent random samples of 100 health-care workers wearing rings and 100 health-care workers not wearing rings are selected. What is the standard deviation of the sampling distribution of the difference in the sample proportions (wear rings minus does not wear rings) of health-care workers having bacteria on one or both hands?

Answer 1

The standard deviation of the sampling distribution of the difference in the sample proportions is approximately 0.040.

Step-by-step explanation:

The standard deviation of the sampling distribution of the difference in the sample proportions can be calculated using the following formula:

√((p1*(1-p1))/n1 + (p2*(1-p2))/n2)

Where p1 and p2 are the sample proportions, and n1 and n2 are the sample sizes. In this case, p1 = 0.31, p2 = 0.27, n1 = 100, and n2 = 100.

Substituting these values into the formula:

√((0.31*(1-0.31))/100 + (0.27*(1-0.27))/100) = 0.040

Therefore, the standard deviation of the sampling distribution of the difference in the sample proportions is approximately 0.040.

Answer 2

C: square root 0.31 (0.69)/100 + 0.27(0.73) /100

Step-by-step explanation:

that's the answer I'm just to lazy to put the explanation but yeah, that's the answer