Question

The results of a mathematics placement exam at two different campuses of Mercy College follow: Campus Sample Size Sample Mean Population Standard Deviation 1 330 33 8 2 310 31 7 What is the alternative hypothesis if we want to test the hypothesis that the mean score on Campus 1 is higher than on Campus 2

Answer 1

Answer:

$z=\frac{(33-31)-0}{\sqrt{\frac{8^2}{330}+\frac{7^2}{310}}}}=3.37$  

$p_v =P(Z>3.37)=1-P(Z$  

Comparing the p value with a significance level for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the mean for the Campus 1 is significantly higher than the mean for the group 2.  

Step-by-step explanation:

Data given

Campus   Sample size     Mean    Population deviation

   1                 330               33                      8

   2                310                31                       7

$\bar X_{1}=33$ represent the mean for sample 1  

$\bar X_{2}=31$ represent the mean for sample 2  

$\sigma_{1}=8$ represent the population standard deviation for 1  

$\sigma_{2}=7$ represent the population standard deviation for 2  

$n_{1}=330$ sample size for the group 1  

$n_{2}=310$ sample size for the group 2  

$\alpha$ Significance level provided  

z would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the mean for Campus 1 is higher than the mean for Campus 2, the system of hypothesis would be:

Null hypothesis:$\mu_{1}-\mu_{2}\leq 0$  

Alternative hypothesis:$\mu_{1} - \mu_{2}> 0$  

We have the population standard deviation's, and the sample sizes are large enough we can apply a z test to compare means, and the statistic is given by:  

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

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

With the info given we can replace in formula (1) like this:  

$z=\frac{(33-31)-0}{\sqrt{\frac{8^2}{330}+\frac{7^2}{310}}}}=3.37$  

P value  

Since is a one right tailed test the p value would be:  

$p_v =P(Z>3.37)=1-P(Z$  

Comparing the p value with a significance level for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the mean for the Campus 1 is significantly higher than the mean for the group 2.