We know, ~ t-distribution with (n-1) d.f.
1. The 95% confidence interval of the population mean is:
[, ], where, = 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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The first photo has the correct numbers and percentage but the second photos allows you to...
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