# A survey of 1,056 tablet owners was conducted. construct the confidence intervals as follows

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A survey of 1,056 tablet owners was conducted. In response to a survey question about shopping, 27% of tablet owners said they use mobile devices for payment, 21% said they use such devices to make social media comments about their purchases, and 9% said they use such devices to retrieve mobile coupons. Complete parts (a) through (d) below.

a. Construct a 90% confidence interval estimate of the population proportion pi
of tablet owners who said they use mobile devices for payment.

b. Construct a 90% confidence interval estimate of the population proportion pi of tablet owners who said they use mobile devices to make social media comments about their purchases.

c. Construct a 90% confidence interval estimate of the population proportion pi of tablet owners who said they use mobile devices to retrieve mobile coupons.

d. You have been asked to update the results of this study. Determine the sample size necessary to estimate, with 90% confidence, the population proportions in (a) through (c) to within +/- 0.05.

I will teach you how to construct the confidence interval using Excel.

First, note that the sample used here is 1056

NOTE THE CONFIDENCE INTERVALS CONSTRUCTED BELOW ARE FOR THE 99% CONFIDENCE INTERVAL

Confidence interval for the proportion of people who are table owners is as follows;-
Apply the formula for the confidence interval.

p_hatch +/- Zc * sqrt(p_hatch * (1 - p_hatch)/n)

The z - score is calculated in Excel as follows;-
=NORM.S.INV(1-0.01/2) = 2.58 (tO find )

the sample proportion is 0.21

0.21 +/- 2.58 * sqrt(0.21 * (1-0.21)/1056)

Margin of error = 0.032

0.21 +/- 0.032, The required confidence interval is (0.178, 0.242)

The above is the confidence interval of the proportion of people who said they use mobile devices to read social comments about their purchases.

The confidence interval of Those who said they use the devices for payments is.

0.27 +/- 2.58 * sqrt(0.27*(1-0.27)/1056)

The confidence interval is 0.27 +/- 0.035

The confidence interval for the problem in a is (0.235, 0.305)

The proportion of people who use the devices to retrieve their mobile coupons is

0.09 +/- 2.58 * sqrt(0.09 * (1 - 0.09)/1056)

The confidence interval is;- 0.09 +/- 0.023, and the confidence interval is (0.067, 0.113).

What if the sample size for 90% and a margin of error of +/- 0.05.

The Z score is;- =NORM.S.INV(1-0.1/2);- =NORM.S.INV(1-0.1/2) = 1.645

the sample size is when p_hatch = 0.27 is, n = 0.27*0.73*(1.645/0.05)^2 = 214 (Sample size rounded to the whole number).

sample size needed when p_hatch is 0.21 is, n = 0.21 * 0.79*(1.645/0.05)^2 = 180 (Round up the value)

The sample size that would be needed in part c is,

n = 0.09 * 0.91 * (1.645/0.05)^2 =  89