Gm535

A. The average (mean) annual income was less than $50,000

Solution:

1. Null hypothesis (Ho): The average(mean) annual income was ≤$50,000

Ho: µ ≥ 50000

2. Alternative (research) hypothesis (Ha): The average (mean) annual income < $50,000

Ha: µ < 50000

3. Test statistic:

As the alternative hypothesis is Ha: µ < 50000, the given test is a one-tailed (lower-tailed) z-test.

One-Sample Z: Income ($1000)

Test of mu = 50 vs. < 50

The assumed standard deviation = 14.64

95% Upper

Variable N Mean StDev SE Mean Bound Z P

Income ($1000) 50 43.74 14.64 2.07 47.15 -3.02 0.001

4. Rejection region:

The critical value for significance level, α =0.05 for a lower-tailed z-test is given as -1.645.

Decision Rule: Reject H0, if z-statistic <-1.645

5. Assumptions:

We shall use Significance Level, α=0.05.

Since the sample size; n > 30, I will use the one-sample Z test for mean to test the given hypothesis.

6. Experiment and calculation of test statistic: (MINITAB OUTPUT)

One-Sample Z: Income ($1000)

The assumed standard deviation = 14.64

Variable N Mean StDev SE Mean 95% CI

Income ($1000) 50 43.74 14.64 2.07 (39.68, 47.80)

Z= (43.74-50)/ (14.64/√50)

Z= -6.26/2.070408655

Z= -3.023557685

Z -3.02 (round)

7. Conclusion:

Since the P-value (0.001) is smaller than the significance level (0.05), we reject the null hypothesis. The p-value involves the probability of rejecting a true null hypothesis.

Thus, at a significance level of 0.05, there is sufficient evidence to support the claim that the average (mean) annual income was less than $50,000. Also, the test statistic of Z= -3.02 does fall in the rejection of z<-1.645; therefore, I would reject the null hypothesis because...