Question

Suppose the following data are selected randomly from a population of normally distributed values.

According to the U.S. Bureau of Labor Statistics, the average weekly earnings of a production worker in July 2011 were $657.49. Suppose a labor researcher wants to test to determine whether this figure is still accurate today. The researcher randomly selects 56 production workers from across the United States and obtains a representative earnings statement for one week from each. The resulting sample average is $670.76. Assuming a population standard deviation of $63.90 and a 10% level of significance, determine whether the mean weekly earnings of a production worker have changed. Appendix A Statistical Tables (Round your answer to 2 decimal places.) The value of the test statistic is and we (reject the null hypothesis/fail to reject the null hypothesis) .

Construct a 95% confidence interval to estimate the population mean.

Appendix A Statistical Tables

(Round the intermediate values to 2 decimal places. Round your answers to 2 decimal places.)

≤ μ ≤

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 657.49
Alternative Hypothesis, Ha: μ ≠ 657.49

Rejection Region
This is two tailed test, for α = 0.1
Critical value of z are -1.645 and 1.645.
Hence reject H0 if z < -1.645 or z > 1.645

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (670.76 - 657.49)/(63.9/sqrt(56))
z = 1.55

P-value Approach
P-value = 0.1211
As P-value >= 0.1, fail to reject null hypothesis.

sample mean, xbar = 670.76
sample standard deviation, σ = 63.9
sample size, n = 56

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

ME = zc * σ/sqrt(n)
ME = 1.96 * 63.9/sqrt(56)
ME = 16.74

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (670.76 - 1.96 * 63.9/sqrt(56) , 670.76 + 1.96 * 63.9/sqrt(56))
CI = (654.02 , 687.5)