Question

From a sample of 37 stolen watches, you find that the mean cash value is $950. The standard deviation for your sample is $90. Test the null hypothesis that your sample came from a population whose mean cash value (Mu) is $1,000 against the alternative hypothesis that Mu is different than $1,000. Use alpha = .05.

Formally state the alternative and the null hypothesis for this test (3 pts)

What is the critical value(s) for this test? (3 pts)

Compute and report the test statistic. (3 pts)

Interpret your decision regarding the null hypothesis (3 pts)

If necessary, compute 95% confidence intervals for the population mean and interpret. (3 pts)

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis: μ = 1000
Alternative Hypothesis: μ ≠ 1000

Rejection Region
This is two tailed test, for α = 0.05 and df = 36
Critical value of t are -2.028 and 2.028.
Hence reject H0 if t < -2.028 or t > 2.028

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (950 - 1000)/(90/sqrt(37))
t = -3.379

P-value Approach
P-value = 0.0018
As P-value < 0.05, reject the null hypothesis.

There is sufficient evidence to conclude that Mu is different than $1,000

CI:
CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (950 - 2.03 * 90/sqrt(37) , 950 + 2.03 * 90/sqrt(37))
CI = (919.96 , 980.04)