Question

1. The Bureau of Economic Analysis in the U.S. Department of Commerce reported that the mean annual income for a resident of North Carolina is $ 18,688 with a Standard Deviation of 15000 (USA Today, August 24, 1995). A researcher for the state of South Carolina wants to test the following hypothesis: where μ is the mean annual income for a resident of South Carolina. The researcher gathers information from a sample of 625 residents of South Carolina and finds a sample mean of 17,076 (=17,076) with a sample standard deviation (s) equal to 15,500.

a. What is the appropriate conclusion pertaining to the hypothesis formulated above? Use a .01 level of significance.

b. Construct a 99% Confidence Interval for the value of the mean to mean of the South Carolina residents’ income(assuming you have no knowledge of the population mean and rely solely on the information from your sample).

c. How would your answer to part a change if you did not know the value of the population standard deviation and your sample size was only 25?.

d. What if the researcher was only concerned that the mean income reported by the Bureau might have intentionally exaggerated the mean income for the State? How would your answer to part a will change? Demonstrate.

In all these problems, show the area under the curve to graphically demonstrate your answers.

Answer 1

a. What is the appropriate conclusion pertaining to the hypothesis formulated above? Use a .01 level of significance.

The Null and Alternative Hypotheses are,

Since the population standard deviation is known, the z-tet is used to test the hypothesis.

The z statistic is obtained using the formula,

The p-value is obtained from the z distribution table for z = -2.687

Since the P-value is less than the significance level = 0.01 at 1% significance level, it can be concluded that the null hypothesis is rejected.

(the plot is obtained in free online software StatKey)

The standard normal distribution plot shows the critical region for significance level = 0.01 for two-tailed hypotheses. we can see the z value lies in the critical region hence the null hypothesis is rejected.

b)

Since we are not known about the population distribution, the confidence interval for the mean is obtained using the formula (using t distribution),

The t critical value is obtained from t distribution table for significance level = 0.10 and degree of freedom = n -1 = 625 - 1 = 624.

c)

Sample size = 25

Since we are not known about the population standard deviation, the t distribution is used to test the hypothesis,

The t statistic is obtained using the formula,

P-value

The p-value is obtained from t distribution table for t = -0.52 and degree of freedom = n - 1 = 25 - 1 = 14. (In excel use function =T.DIST.2T(0.52,24))

Conclusion

Since the P-value is greater than the significance level = 0.01 at 1% significance level, the null hypothesis is not rejected.

The st distribution plot shows the critical region for significance level = 0.01 for two-tailed hypotheses. we can see the t value does not lie in the critical region hence the null hypothesis is not rejected.

d)

Since we are not concerned with the population standard deviation, now using the t distribution,

The t statistic is obtained using the formula,

P-value

The p-value is obtained from t distribution table for t = -2.6 and degree of freedom = n - 1 = 625 - 1 = 624. (In excel use function =T.DIST.2T(2.6,624))

Conclusion

Since the P-value is less than the significance level = 0.01 at 1% significance level, the null hypothesis is rejected.

Hence we can conclude that mean income reported by the Bureau might have intentionally exaggerated the mean income for the State