Question

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.

Answer 1

Answer:

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. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = $\frac{\sum X}{n}$ = $\frac{4.4+5.2+ 6.4+6.8+7.1+ 7.3+ 8.3+ 8.4}{8}$ = 6.74

             s = sample standard deviation = $\frac{\sum(X-\bar X)^{2} }{n-1}$ = 1.4

             n = sample of values = 8

             $\mu$ = population mean

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

P(-2.365 < $t_7$ < 2.365) = 0.95

P(-2.365 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 2.365) = 0.95

P( $-2.365 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.365 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X -2.365 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X +2.365 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X -2.365 \times {\frac{s}{\sqrt{n} } }$ , $\bar X +2.365 \times {\frac{s}{\sqrt{n} } }$ ]

                                                = [ $6.74 -2.365 \times {\frac{1.4}{\sqrt{8} } }$ , $6.74 +2.365 \times {\frac{1.4}{\sqrt{8} } }$ ]

                                                = [5.57 , 7.91]

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

Answer 2

Answer:

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] ± $t_{n-1;1-\alpha /2}$* (S/√n)

$t_{n-1;1-\alpha /2}= t_{7;0.975}= 2.365$

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!