Question

A statistics teacher wants to see if there is any difference in the abilities of students enrolled in statistics today and those enrolled five years ago. A sample of final examination scores from students enrolled today and from students enrolled five years ago was taken. You are given the following information.
Today Five Years Ago
82 88
σ2 112.5 54
n 45 36
Refer to Exhibit 10-3. The p-value for the difference between the two population means is
a) 0013
b) 0026
c) 4987
d) 9987

Answer 1

Answer:

$p_v =2*P(Z<-3)=0.026997$

Step-by-step explanation:

Data given:

                    Today    Five Years Ago

Mean  82  88

Variance        112.5  54

Sample Size 45  36

Data given and notation

$\bar X_{today}=82$ represent the mean for the sample today

$\bar X_{five years ago}=88$ represent the mean for the sample five years ago

$\sigma_{today}=10.607$ represent the population standard deviation for the sample today

$\sigma_{five years ago}=7.348$ represent the population standard deviation for the sample five years ago

$n_{today}=45$ sample size for the group today

$n_{five years ago}=36$ sample size for the group five years ago

t would represent the statistic (variable of interest)

$\alpha$ significance level

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the means for the two groups are different, the system of hypothesis would be:

Null hypothesis:$\mu_{today}=\mu_{five years ago}$

Alternative hypothesis:$\mu_{today} \neq \mu_{five years ago}$

Since we know the population deviations for each group, for this case is better apply a z test to compare means, and the statistic is given by:

$z=\frac{\bar X_{today}-\bar X_{five years ago}}{\sqrt{\frac{\sigma^2_{today}}{n_{today}}+\frac{\sigma^2_{five years ago}}{n_{five years ago}}}}$ (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.

Determine the critical value(s).

Based on the significance level assumed $\alpha=0.01$ and $\alpha/2=0.005$ we can find the critical values with the normal standard distribution, we are looking for values that accumulates 0.005 of the area on each tail on the normal distribution.

For this case the two values are $z_{\alpha/2}=-2.58$ and $z_{1-\alpha/2}=2.58$

Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

$z=\frac{82-88}{\sqrt{\frac{10.607^2}{45}+\frac{7.348^2}{36}}}}=-3$  

What is the p-value for this hypothesis test?

Since is a bilateral test the p value would be:

$p_v =2*P(Z<-3)=0.026997$