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

Answer:

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

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:

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

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

Replacing we got:

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

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