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.
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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