Question

The mean class grade for all statistic students from the 2016-2017 was 85% with a standard deviation of 7%. We want to know if the current class of statistic students is significant different from last year's population. The current mean class grade is 88% with a standard deviation of 5% and there are 30 students in the class. Use α = .01.

Answer 1

Answer:

$z=\frac{85-88}{\sqrt{\frac{7^2}{30}+\frac{5^2}{30}}}}=-1.91$  

$p_v =2*P(Z$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can't conclude that the two means are significantly different at 1% of significance.

Step-by-step explanation:

Data given and notation

$\bar X_{A}=85$ represent the mean for the sample of 2016-2017

$\bar X_{B}=88$ represent the mean for the sample of the last year

$\sigma_{A}=7$ represent the population standard deviation for the sample of 2016-2017

$\sigma_{B}=5$ represent the population standard deviation for the sample of last year

$n_{A}=30$ sample size selected for the sample of 2016-2017

$n_{B}=30$ sample size selected for the last year

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the two means are equal or no, the system of hypothesis would be:

Null hypothesis:$\mu_{A} = \mu_{B}$

Alternative hypothesis:$\mu_{A} \neq \mu_{B}$

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

$z=\frac{\bar X_{A}-\bar X_{B}}{\sqrt{\frac{\sigma^2_{A}}{n_{A}}+\frac{\sigma^2_{B}}{n_{B}}}}$ (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".

Calculate the statistic

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

$z=\frac{85-88}{\sqrt{\frac{7^2}{30}+\frac{5^2}{30}}}}=-1.91$  

P-value

Since is a two tailed test the p value would be:

$p_v =2*P(Z$

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can't conclude that the two means are significantly different at 1% of significance.