Question

Independent random samples taken on two university campuses revealed the following information concerning the average amount of money spent on textbooks during the fall semester.University A University BSample Size 50 40Average Purchase $280 $250Standard Deviation $20 $23At * = .05, test to determine if, on the average, students at University A spend more on textbooks then the students at University B

Answer 1

Answer:

$p_v =P(t_{88}>6.960) =2.91x10^{-10}$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group A is significantly lower than the mean for the group B.  

Step-by-step explanation:

The statistic to test the hypothesis is given by this formula:

$t=\frac{(\bar X_A-\bar X_B)-(\mu_{A}-\mu_B)}{\sqrt{\frac{s_A^2}{n_A}}+\frac{s_B^2}{n_B}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom

The system of hypothesis on this case are:

Null hypothesis: $\mu_A \leq \mu_B$

Alternative hypothesis: $\mu_A >\mu_B$

Or equivalently:

Null hypothesis: $\mu_A - \mu_B \leq 0$

Alternative hypothesis: $\mu_A -\mu_B>0$

Our notation on this case :

$n_A =50$ represent the sample size for group A

$n_B =40$ represent the sample size for group B

$\bar X_A =280$ represent the sample mean for the group A

$\bar X_B =250$ represent the sample mean for the group B

$s_A=20$ represent the sample standard deviation for group A

$s_B=23$ represent the sample standard deviation for group B

And now we can calculate the statistic:

$t=\frac{(280 -250)-(0)}{\sqrt{\frac{20^2}{50}}+\frac{23^2}{40}}=6.511$

Now we can calculate the degrees of freedom given by:

$df=50+40-2=88$

And now we can calculate the p value using the alternative hypothesis:

$p_v =P(t_{88}>6.960) =2.91x10^{-10}$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group A is significantly lower than the mean for the group B.  

Answer 2

Answer:

Step-by-step explanation:

Download docx