Question

# A researcher is interested in finding a 98% confidence interval for the mean number minutes students...

A researcher is interested in finding a 98% confidence interval for the mean number minutes students are concentrating on their professor during a one hour statistics lecture. The study included 147 students who averaged 34.2 minutes concentrating on their professor during the hour lecture. The standard deviation was 10.6 minutes.

a. To compute the confidence interval use a ? t z  distribution.

b. With 98% confidence the population mean minutes of concentration is between  and   minutes.

c. If many groups of 147 randomly selected members are studied, then a different confidence interval would be produced from each group. About  percent of these confidence intervals will contain the true population mean minutes of concentration and about  percent will not contain the true population mean minutes of concentration.

Given,

sample n = 147

mean = 34.2

standard deviation = 10.6

a)

Here the confidence interval use a t distribution

b)

degree of freedom = n - 1 = 147 - 1 = 146

alpha = 0.02

t(alpha/2 ,df) = t(0.02/2 , 146) = 2.35216 ~ 2.35

CI = xbar +/- t*s/sqrt(n)

substitute values

= 34.2 +/- 2.35*10.6/sqrt(147)

= 34.2 +/- 2.0545

= (32.1455 , 36.2545)

= (32.15 , 36.25)

c)

About 98 percent of these confidence intervals will contain the true population mean minutes of concentration and about 2 percent will not contain the true population mean minutes of concentration.

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