a. the average (mean) annual income was less than $50,000

Null and Alternative Hypothesis

H0: mu= 50 (in thousands)

Ha: mu<50 (in thousands)

Level of Significance

Level of Significance = .05

Test Statistic, Critical Value, and Decision Rule

Since alpha = .05, z<-1.645, which is lower tailed

Rejection region is, z<-1.645

Calculate test statistic, x-bar=43.74 and s=14.64

Z=(43.74-50)/2.070=-3.024 2.070 is calculated by: s/sq-root of n

Decision Rule: The calculated test statistic of -3.024 does fall in the rejection region of z<-1.645, therefore I would reject the null and say there is sufficient evidence to indicate mu<50.

Interpretation of Results and Conclusion

p-value= .001

.001<.05

Because the p-value of .001 is less than the significance level of .05, I will reject the null hypothesis at 5% level.

95% CI=(39.68, 47.80)- I am 95% confident that the true mean income lies between $39,680 and $47,800. Minitab Output: One-Sample Z

Test of mu = 50 vs < 50

The assumed standard deviation = 14.64

95% Upper N Mean SE Mean Bound Z P 50 43.74 2.07 47.15 -3.02 0.001

b. the true population proportion of customers who live in an urban area exceeds 40%

22 people of the 50 surveyed live in an Urban community, which is 44%. My point estimate is .44.

Null and Alternative Hypothesis

H0: p=.40

Ha: p>.40

Level of Significance

Level of Significance= .05

Test Statistic, Critical Value, and Decision Rule

Since alpha= .05, z>1.645, which is upper tailed

Rejection region is, z>1.645

To conduct a large sample z-test, I must first determine if the sample size is large enough. nPo= 50(.40)= 20 and 50(1-.40)=30

Both are larger than 15, so we conclude that the sample size is large enough to conduct the large sample z test.

Z=(.44-.40)/.06928=.5774 .06928 is calculated by sq-root

References: Benson, P. G., McClave, J. T., & Sincich, T. (2011). Statistics for Business and Economics (11th ed.). Boston, MA: Prentice Hall.