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 equals 5.n=5.

1,2,3,4 and 17

In the given data, replace the value 17 with 5 and recalculate the confidence interval. Using these results, describe the effect of an outlier (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.

Answer #1

Here, mean = 3 , std.dev = 1.5811

n = 5

at 95% confidence interval the t value is ,

df =n -1 = 5 - 1 =4

alpha = 1- 0.95 = 0.05

alpha/2 = 0.05/2 = 0.025

t(alpha/2,df) = t (0.025,4) = 2.7764

Margin of error = t *(sigma/sqrt(n))

= 2.7764 *(1.5811/sqrt(5))

= 1.9632

At 95% confidence interval the mean is,

mean -E < mu < mean +E

3 - 1.9632 < mu < 3 + 1.9632

1.0368 < mu < 4.9632

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