Respuesta :
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] ± [tex]t_{n-1;1-\alpha /2}[/tex]* (S/√n)
[tex]t_{n-1;1-\alpha /2}= t_{7;0.975}= 2.365[/tex]
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!
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. = [tex]\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = [tex]\frac{\sum X}{n}[/tex] = [tex]\frac{4.4+5.2+ 6.4+6.8+7.1+ 7.3+ 8.3+ 8.4}{8}[/tex] = 6.74
s = sample standard deviation = [tex]\frac{\sum(X-\bar X)^{2} }{n-1}[/tex] = 1.4
n = sample of values = 8
[tex]\mu[/tex] = population mean
So, 95% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-2.365 < [tex]t_7[/tex] < 2.365) = 0.95
P(-2.365 < [tex]\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.365) = 0.95
P( [tex]-2.365 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X - \mu}[/tex] < [tex]2.365 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
P( [tex]\bar X -2.365 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X +2.365 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X -2.365 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X +2.365 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]6.74 -2.365 \times {\frac{1.4}{\sqrt{8} } }[/tex] , [tex]6.74 +2.365 \times {\frac{1.4}{\sqrt{8} } }[/tex] ]
= [5.57 , 7.91]
Therefore, 95% confidence interval for the mean amount of time that students study per week is [5.57 , 7.91].
