Question

A study is done by a community group in two neighboring colleges to determine which one graduates students with more math classes. College A samples 11 graduates and finds the mean is 4 math classes with a standard deviation of 1.5 math classes. College B samples 9 graduates and finds the mean is 3.5 math classes with a standard deviation of 1 math class. The community group believes that a student who graduates from college A has taken more math classes, on the average. Test at a 10% significance level. Assume the requirements for a valid hypothesis test are satisfied.

Answer 1

To determine if there is a significant difference in the average number of math classes taken by graduates of College A and College B, a hypothesis test can be conducted using a t-test.

Step-by-step explanation:

To determine whether there is a significant difference in the average number of math classes taken by graduates of College A and College B, a hypothesis test can be conducted. The null hypothesis (H0) states that there is no difference between the means of the two populations, while the alternative hypothesis (Ha) states that the mean of College A is higher than the mean of College B.

The test statistic used for comparing means of two independent samples is the t-test. By plugging in the sample means, sample standard deviations, and sample sizes into the formula for the t-test, we can calculate the t-value.

Looking up the critical value of the t-distribution for a 10% significance level with degrees of freedom equal to the sum of the sample sizes minus 2, we can compare the t-value to the critical value. If the t-value is greater than the critical value, we reject the null hypothesis and conclude that there is a significant difference in the average number of math classes taken by graduates of the two colleges.

Answer 2

`t=\frac{(4-3.5)-0}{\sqrt{(1.5^2)/(11)+(1^2)/(9)}}}=0.890`

The p value for this case would be:

`p_v =P(t_(18)>0.890)=0.193`

The p value is higher than the significance level so then we can conclude that we can FAIL to reject the null hypothesis and then the true mean for group A is not significantly higher than the mean for B

Step-by-step explanation:

Information given

`\bar X_(1)=4` represent the mean for sample A

`\bar X_(2)=3.5` represent the mean for sample B

`s_(1)=1.5` represent the sample standard deviation for A

`s_(2)=1` represent the sample standard deviation for B

`n_(1)=11` sample size for the group A

`n_(2)=9` sample size for the group B

`\alpha=0.1` Significance level provided

t would represent the statistic

Hypothesis to test

We want to verify if the student who graduates from college A has taken more math classes, on the average, the system of hypothesis would be:

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

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

The statistic is given by:

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

And the degrees of freedom are given by
`df=n_1 +n_2 -2=11+9-2=18`

Replacing we got:

`t=\frac{(4-3.5)-0}{\sqrt{(1.5^2)/(11)+(1^2)/(9)}}}=0.890`

The p value for this case would be:

`p_v =P(t_(18)>0.890)=0.193`

The p value is higher than the significance level so then we can conclude that we can FAIL to reject the null hypothesis and then the true mean for group A is not significantly higher than the mean for B