According to a 1997 survey of public participation in the arts, Americans who us

Question

According to a 1997 survey of public participation in the arts, Americans who used personal computers at home during their free time averaged 5.2 hours per week on their computers. Somebody doubts that the mean is more than 5.2 hours this year. A recent random sample of 50 Americans who use computers during their free time showed that the mean time these individuals spent on computers is 6.1 hours with a standard deviation of 2.3 hours. The significance level is 5%.

Part I) critical value method

a) Set up the suitable null hypothesis and alternative hypothesis.

b) Find the value of the test statistic. What distribution will we use in this case?

c) Identify the rejection region.

d) State your conclusion for this test of hypothesis.

e)What is type I error in this problem?

Explanation / Answer

here null hypothesis: mean <=5.2

alternate hypothesis: mean >5.2

b) as we do not know std deviaiton of population we will use t distribution. How-ever as sample size is high z distribution will approximately match it.

std error =std deviation/(|n)1/2 =0.3253

hence test stat t =(X-mean)/std error =(6.1-5.2)/0.3253=2.7667

c) for 0.05 level critical value =1.6765

d) as our test stat falls into rejection region; we can reject null hypothesis.

E) type I error is rejecting null hypothesis that mean is equal to 5.2 ; even though it is true,

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.