We are interested in the amount that students study per week. Suppose you collected the following data in hours {4.4, 5.2, 6.4, 6.8, 7.1, 7.3, 8.3, 8.4}. Construct a 95% confidence interval for the mean amount of time that students study per week.

Step-by-step explanation:

Hello!

The objective is to estimate the average time a student studies per week.

A sample of 8 students was taken and the time they spent studying in one week was recorded.

4.4, 5.2, 6.4, 6.8, 7.1, 7.3, 8.3, 8.4

n= 8

X[bar]= ∑X/n= 53.9/8= 6.7375 ≅ 6.74

S²= 1/(n-1)*[∑X²-(∑X)²/n]= 1/7*[376.75-(53.9²)/8]= 1.94

S= 1.39

Assuming that the variable "weekly time a student spends studying" has a normal distribution, since the sample is small, the statistic to use to perform the estimation is the student's t, the formula for the interval is:

X[bar] ± * (S/√n)

6.74 ± 2.365 * (1.36/√8)

[5.6;7.88]

Using a confidence level of 95% you'd expect that the average time a student spends studying per week is contained by the interval [5.6;7.88]

I hope this helps!

95% confidence interval for the mean amount of time that students study per week is [5.57 , 7.91].

Step-by-step explanation:

We are given with the following data in hours;

4.4, 5.2, 6.4, 6.8, 7.1, 7.3, 8.3, 8.4

So, the pivotal quantity for 95% confidence interval for the mean amount of time that students study per week is given by;

           P.Q. = ~

where, = = = 6.74

             s = sample standard deviation = = 1.4

             n = sample of values = 8

              = population mean

So, 95% confidence interval for the population mean, is ;

P(-2.365 < < 2.365) = 0.95

P(-2.365 < < 2.365) = 0.95

P( < < ) = 0.95

P( < < ) = 0.95

95% confidence interval for = [ , ]

                                                = [ , ]

                                                = [5.57 , 7.91]

Therefore, 95% confidence interval for the mean amount of time that students study per week is [5.57 , 7.91].

im not sure

step-by-step explanation: