Question

2) Given *n* = 16, *x**=12.37*, and
*s* = 1.05 from a normally distributed population, construct
a 99% confidence interval about σ.

3) A poll of 100 randomly selected 50 year old people were asked in how many years were they expecting to retire. The mean was found to be 7.2 years with a standard deviation of 3.75 years. Test the claim that the mean for all 50 year old people is less than 7.5 years assuming a 0.05 significance level.

Answer #1

2)

Here s = 1.05 and n = 16

df = 16 - 1 = 15

α = 1 - 0.99 = 0.01

The critical values for α = 0.01 and df = 15 are Χ^2(1-α/2,n-1) =
4.601 and Χ^2(α/2,n-1) = 32.801

CI = (sqrt(15*1.05^2/32.801) , sqrt(15*1.05^2/4.601))

CI = (0.71 , 1.90)

3)

Below are the null and alternative Hypothesis,

Null Hypothesis, H0: μ = 7.5

Alternative Hypothesis, Ha: μ < 7.5

Rejection Region

This is left tailed test, for α = 0.05 and df = 99

Critical value of t is -1.66.

Hence reject H0 if t < -1.66

Test statistic,

t = (xbar - mu)/(s/sqrt(n))

t = (7.2 - 7.5)/(3.75/sqrt(100))

t = -0.8

P-value Approach

P-value = 0.2128

As P-value >= 0.05, fail to reject null hypothesis.

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