Question

a) Suppose each of the following confidence intervals were calculated using the same sample. Circle the interval that is the lower confidence level. State the sample proportion that was used to construct these intervals.

(0.25, 0.35) (0.24, 0.36)

b) Suppose you want a confidence interval to be more narrow. What two things can you do to achieve this?

c) For a given sample, which would you prefer, a wider interval or a more narrow one? Why?

d) Suppose your sample proportion is 0.17, with n= 100. How many "success" were in your sample?

e) Suppose you have a sample with n= 27.9 of your sample values are found to have the characteristic of interest. Are the 4 conditions required to do a confidence interval met?
f) True or False: a 95% confidence interval for p always contains p.

Answer 1

(A) we know that proportion = (lower limit+upper limit)/2

Proportion for first confidence interval = (0.35+0.25)/2= 0.60/2 = 0.30

So, margin of error = upper limit - proportion = 0.35-0.30 = 0.05

Proportion for second confidence interval = (0.36+0.24)/2= 0.60/2 = 0.30

So, margin of error = upper limit - proportion = 0.36-0.30 = 0.06

we know that smaller margin of error means smaller confidence level and higher margin of error means higher confidence level. So, confidence interval one has smaller confidence level.

(B) we can decrease the confidence interval and increase the sample size n, both these steps will reduce the margin of error which will narrow confidence interval

(C) I will prefer wider confidence interval because wider confidence interval will include more data values as compared to narrow confidence interval

(D) Number of success = n*p

where n = 100 and p = 0.17

so, number of success = 100*0.17 = 17

(E) No, because the sample size n can never be a decimal number. Therefore, conditions are not met.

(F) Yes, it is true because the p value is always present in a confidence interval irrespective of confidence level.