6. Newspaper Reading Times A surve years ago found that the average time a readi

Question

6. Newspaper Reading Times A surve years ago found that the average time a reading the local daily newspaper was 10.8e The standard deviation of the population To see whether the average time had changed si newspaper's format was revised, the newspa surveyed 36 individuals. The average time that t 36 people spent reading the paper was 12.2 minutes ea eronl d ch since the ewspaper edito At = 0.02, is there a change in the average time an individual spends reading the newspaper? Find the 98% confidence interval of the mean. Do the results agree? Explain.

Explanation / Answer

here null hypothesis:Ho: =10.8

alternate hypothesis:Ha: 10.8

for 98% CI crtiical value of z =2.326

Decision rule : reject Ho if test statistic z>2.326 ot z<-3.26

std error of mean =std deviaiton/(n)1/2 =3/(36)1/2 =0.5

hence test statistic z=(X-mean)/std error =(12.2-10.8)/0.5 =2.8

as test statistic is in critical region we reject null hypothesis . we have suffiient evidence to conclude that there is a change in average time an individual spends reading the newspaper

Related Questions

Navigate

• Browse (All)
• Browse 6
• Subjects

---

---

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