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A 2015 Gallup poll of 1,627 adults found that only 22% felt fully engaged with their mortgage provider.

What is the sample size?




The margin of error for this data set is 2.5%. What is the 95% confidence interval using only 1 decimal?

User Thiyagu
by
3.2k points

2 Answers

2 votes

Final answer:

The sample size for the Gallup poll is 1,627 adults. For the 95% confidence interval with a 2.5% margin of error and an engagement rate of 22%, the interval is (19.5%, 24.5%).

Step-by-step explanation:

The sample size mentioned in the Gallup poll is 1,627 adults. To calculate the 95% confidence interval for the estimated proportion of adults who felt fully engaged with their mortgage provider, we use the provided percentage (22%) and the margin of error (2.5%).

The confidence interval formula is:

Confidence Interval = Estimate ± Margin of Error

In this case, the point estimate is 22% (or 0.22 as a decimal). The margin of error (MOE) is 2.5% (or 0.025 as a decimal).

To find the confidence interval, we subtract and add the margin of error from the point estimate:

Lower limit = 0.22 - 0.025 = 0.195

Upper limit = 0.22 + 0.025 = 0.245

Therefore, the 95% confidence interval in decimal form is (0.195, 0.245). Converting to percentages and rounding to one decimal place, we obtain:

95% Confidence Interval: (19.5%, 24.5%)

User Lante Dellarovere
by
3.3k points
6 votes

Answer:
(20.0\%,24.0\%)

Step-by-step explanation:

Given, A 2015 Gallup poll of 1,627 adults found that only 22% felt fully engaged with their mortgage provider.

Here , 1,627 adults are determining the sample.

Thus, sample size : n = 1627

Also, the sample proportion of adults felt fully engaged with their mortgage provider are
\hat{p}=22\%=0.22

for 95% confidence level , critical z-value =1.96

Then , the 95% confidence interval would be :-


\hat{p}\pm z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}

Substituting values , we get


0.22\pm 1.96(\sqrt{(0.22(1-0.22))/(1627)})\\\\=0.22\pm 1.96\sqrt{(0.22*0.78)/(1627)}\\\\\approx0.22\pm0.0201\\\\=(0.22-0.0201,\ 0.22+0.0201)\\\\=(0.1999,\ 0.2401)\approx(20.0\%,24.0\%)

Hence, the required 95% confidence interval using only 1 decimal :
(20.0\%,24.0\%) .