Question

A survey was conducted at two colleges. 500 students at College A participated in the study. The results indicated that on average, the students spent 15 hours per week doing online assignments and its standard deviation was 5 hours. At College B, 400 students participated in the study. The average hours they worked for online assignments was 20 with a standard deviation of 4 hours. Please test whether there is a true difference in the time students spent for online assignments between the two colleges (using a confidence level of 99%).

Answer 1

`t=\frac{(15-20)-0}{\sqrt{(5^2)/(500)+(4^2)/(400)}}}=-16.67`

The p value would be given by:

`p_v =2*P(t_(898)<-16.67) \approx 0`

Since the p value is a very low value we have enough evidence to reject the null hypothesis and we can conclude that we have significant difference in the means of time spent for online assignments between the two colleges

Explanation:

Data given

`\bar X_(1)=15` represent the mean for sample A

`\bar X_(2)=20` represent the mean for sample B

`s_(1)=5` represent the sample standard deviation for A

`s_(f)=4` represent the sample standard deviation for B

`n_(1)=500` sample size for the group A

`n_(2)=400` sample size for the group B

`\alpha=0.01` Significance level provided

t would represent the statistic

System of hypothesis

The system of hypothesis is the true difference in the time students spent for online assignments between the two colleges, the system of hypothesis would be:

Null hypothesis:
`\mu_(1)-\mu_(2)=0`

Alternative hypothesis:
`\mu_(1) - \mu_(2)\\eq 0`

Since we don't know the deviations the statistic is given by:

`t=\frac{(\bar X_(1)-\bar X_(2))-\Delta}{\sqrt{(\sigma^2_(1))/(n_(1))+(\sigma^2_(2))/(n_(2))}}` (1)

The degrees of freedom are given by
`df=n_1 +n_2 -2=500+400-2=898`

Replacing the info given we got:

`t=\frac{(15-20)-0}{\sqrt{(5^2)/(500)+(4^2)/(400)}}}=-16.67`

The p value would be given by:

`p_v =2*P(t_(898)<-16.67) \approx 0`

Since the p value is a very low value we have enough evidence to reject the null hypothesis and we can conclude that we have significant difference in the means of time spent for online assignments between the two colleges