(Answers are given but could you please work through the problemand show me the steps?)

Chapter 9

Understand the difference between type I and type II errors.

Establish null and alternative hypotheses

The average U.S. daily internet use at home is two hoursand twenty minutes. A sample of 64 homes in Soddy-Daisyshowed an average usage of two hours and 50minutes with a standard deviation of 80minutes. We are interested in determining whether or notthe average usage in Soddy-Daisy is significantlydifferent from the U.S. average.

a.	State the null and alternative hypotheses to be tested.
b.	Compute the test statistic. (sample standard deviation isgiven)
c.	The null hypothesis is to be tested using α = .05.Determine the critical value(s) for this test.
d.	What do you conclude?

ANSWER: Â

a. H0: μ = 140 minutes Ha: μ ≠ 140 minutes b. Test statistic t = 3 c. t.025 = 1.998 and -t.025= -1.998 d. Reject H0 and conclude that the averageusage in Soddy-Daisy is significantly different from the nationalaverage of 140 minutes.

The Bureau of Labor Statistics reported that the average yearlyincome of dentists in the year 2012 was $110,000. A sample of 81dentists, which was taken in 2013, showed an average yearly incomeof $120,000. Assume the standard deviation of thepopulation of dentists’s incomes in 2012 is$36,000.

a.	We want to test and determine if there has been a significantincrease in the average yearly income of dentists.Provide the null and the alternative hypotheses.
b.	Compute the test statistic.
c.	Determine the p-value; and at a 5% level ofsignificance, test the hypotheses.

ANSWER: Â

a. H0: μ ≤ $110,000 Ha: μ > $110,000 b. z = 2.5 c. p-value = .0062 Since the p-value = .0062 < .05, rejectH0. Therefore, it can be concluded that therehas been a significant increase.