Question

A sample of size n=100 is drawn from a normal population whose standard deviation is o=9.4. The sample mean is x=42.53.

Part 1 of 2

(a) Construct a 80% confidence interval for u. Round the answer to
at least two decimal places.

An 80% confidence interval for the mean is __<u<__.

Part 2 of 2

(b) If the population were not approximately normal, would the
confidence interval constructed in part (a) be valid Explain.

The confidence interval constructed in part (a) (would/wouldnt) be valid since the sample size (is/isnt) large.

Answer #1

1)

sample mean, xbar = 42.53

sample standard deviation, σ = 9.4

sample size, n = 100

Given CI level is 80%, hence α = 1 - 0.8 = 0.2

α/2 = 0.2/2 = 0.1, Zc = Z(α/2) = 1.2816

ME = zc * σ/sqrt(n)

ME = 1.2816 * 9.4/sqrt(100)

ME = 1.2

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))

CI = (42.53 - 1.2816 * 9.4/sqrt(100) , 42.53 + 1.2816 *
9.4/sqrt(100))

CI = (41.33 , 43.73)

41.33 < mu < 43.73

2)

The confidence interval constructed in part (a) (wouldnt) be valid since the sample size (is) large.

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