A researcher is interested in finding a 90% confidence interval for the mean number minutes students are concentrating on their professor during a one hour statistics lecture. The study included 114 students who averaged 42.5 minutes concentrating on their professor during the hour lecture. The standard deviation was 11.6 minutes. a. The sampling distribution follows a distribution.
b. With 95% confidence the population mean minutes of concentration is between and minutes.
c. If many groups of 150 randomly selected students 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 number of minutes of concentration and about percent will not contain the true population mean number of minutes of concentration.

Step-by-step explanation:

Given that,

Point estimate = sample mean = \bar x = 32.9

sample standard deviation = s = 12.1

sample size = n = 114

Degrees of freedom = df = n - 1 = 114 - 1 = 113

a) t distribution,

b) At 90% confidence level

\alpha = 1 - 90%

\alpha =1 - 0.90 =0.10

\alpha/2 = 0.05

t\alpha/2,df = t0.05,113 = 1.658

Margin of error = E = t\alpha/2,df * (s /\sqrtn)

= 1.658 * (12.1 / \sqrt 114 )

Margin of error = E = 1.879

The 90% confidence interval estimate of the population mean is,

T  ± E  

= 32.9 ± 1.879

= ( 31.021, 34.779 )

With 90% confidence the population mean minutes of concentration is between 31.021 and 34.779 minutes.

c)  If many groups of 114 randomly selected members are studied, then a different confidence interval would be produced from each group. About 90 percent of these confidence intervals will contain the true population mean minutes of concentration and about 10 percent will not contain the true population mean minutes of concentration.

a. the are of the rectangle is 18.25

b. 18.25 times two is 36.50

rae me : d

its an acute right triangle

step-by-step explanation: