Question

Assuming that the population is normally distributed, construct a 95% confidence interval for the population mean, based on the following sample size of n = 7 1, 2, 3, 4, 5, 6, and 162 In the given data, replace the value 16 with 7 and recalculate the confidence interval. Using these results, describe the effect of an outler (that is an extreme value) on the confidence interval, in general Find a 95% confidence interval for the population mean, using the formula or technology sus (Round to two decimal places as needed.)
The first photo has the correct numbers and percentage but the second photos allows you to see all three questions that will be asked if you have any questions let me know

it 95%

16 & 7

Answer 1

We know, $\frac{\overline{X} -\mu}{s/\sqrt{n}}$ ~ t-distribution with (n-1) d.f.

1. The 95% confidence interval of the population mean is:

[
X - 10.995,0-
,
X + to.375,12-17
], where, $\overline{X}$ = 5.2857, s = 5.0238, n = 7

Thus, 95% C.I. = [5.2857 - 4.6464, 5.2857 + 4.6464] = [0.6393, 9.9321]

2. If we replace 16 with 7, then the 95% C.I. = [4 - 1.9979, 4 + 1.9979] = [2.0021, 5.9979]

So, the presence of outliers widens the confidence interval. Option (B) is the correct choice.

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