Please Answer Fast But make sure there right and answer like 1. 2. 3. 4. 5.
2 answers:
You might be interested in
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] ± * (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!