A class survey in a large class for first-year college students asked, \"About h

Question

A class survey in a large class for first-year college students asked, "About how many minutes do you study on a typical weeknight?" The mean response of the randomly selected 463 students was x bar = 118 minutes. Suppose that we know that the study time follows a Normal distribution with standard deviation sigma = 65 minutes in the population of all first-year students at this university. Use the survey result to give a 99% confidence interval for the mean study time of all first year students. Solve: Margin of error= The 99% confidence interval:

Explanation / Answer

mean=118, standard deviation=sd=65, n=463

(100-alpha)% confidence interval for sample mean=mean±z(alpha/2)*sd/sqrt(n)

here alpha=1%

z(0.005)=2.5758

99% confidence interval for sample mean=mean±z(.01/2)*sd/sqrt(n)

=118± 2.5758*65/sqrt(463)=118±7.78=(110.22,125.78)

margin of error=z(alpha/2)*sd/sqrt(n)=2.5758*65/sqrt(463)=7.78

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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