Answer:
Step-by-step explanation:
Given that :
Mean = 7.8
Standard deviation = 0.5
sample size = 30
Sample mean = 7.3 5.4772
The null and the alternative hypothesis is as follows;
![\mathbf{ H_o: \mu \geq 7.8}](https://tex.z-dn.net/?f=%5Cmathbf%7B%20H_o%3A%20%5Cmu%20%20%5Cgeq%20%207.8%7D)
![\mathbf{ H_1: \mu < 7.8}](https://tex.z-dn.net/?f=%5Cmathbf%7B%20H_1%3A%20%5Cmu%20%20%3C%20%207.8%7D)
The test statistics can be computed as :
![z = \dfrac{X- \mu}{\dfrac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cdfrac%7BX-%20%5Cmu%7D%7B%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
![z = \dfrac{7.3- 7.8}{\dfrac{0.5}{\sqrt{30}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cdfrac%7B7.3-%207.8%7D%7B%5Cdfrac%7B0.5%7D%7B%5Csqrt%7B30%7D%7D%7D)
![z = \dfrac{-0.5}{\dfrac{0.5}{5.4772}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cdfrac%7B-0.5%7D%7B%5Cdfrac%7B0.5%7D%7B5.4772%7D%7D)
![z = - 5.4772](https://tex.z-dn.net/?f=z%20%3D%20-%205.4772)
The p-value at 0.05 significance level is:
p-value = 1- P( Z < -5.4772)
p value = 0.00001
Decision Rule:
The decision rule is to reject the null hypothesis if p value is less than 0.05
Conclusion:
At the 0.05 significance level, there is sufficient information to reject the null hypothesis. Therefore ,we conclude that college students watch fewer movies a month than high school students.
Answer:
5
Step-by-step explanation:
1,2,3,4,5,6
Hshsdusisidjdjfjejdixhdbsisoxnd she
If you simplify, it equals 10.82e.