Question

4. Suppose you would like to construct a confidence interval for the population mean of a particular variable. With that in mind, you take a random sample from the population, and obtain the values shown below. a. Please construct a 84% confidence interval for the population mean using your sample of data. b. Your friend claims that the population mean equals 92. With your sample of data, please use the t-stat method to test this hypothesis using a significance level of 4%. Do you reject or fail to reject your friend’s hypothesis?

83 98 121 72 49 52 90 102 44 112 78 48 63

Answer 1

A)
sample mean, xbar = 77.85
sample standard deviation, s = 25.7936
sample size, n = 13
degrees of freedom, df = n - 1 = 12

Given CI level is 84%, hence α = 1 - 0.84 = 0.16
α/2 = 0.16/2 = 0.08, tc = t(α/2, df) = 1.498

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (77.85 - 1.498 * 25.7936/sqrt(13) , 77.85 + 1.498 * 25.7936/sqrt(13))
CI = (67.13 , 88.57)

B)
Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 92
Alternative Hypothesis, Ha: μ ≠ 92

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (77.85 - 92)/(25.7936/sqrt(13))
t = -1.978

P-value Approach
P-value = 0.0714
As P-value >= 0.04, fail to reject null hypothesis.

There is not sufficient evidence to conclude that the mean is different than 92.