Question

A University of Florida economist conducted a study of Virginia elementary school lunch menus During the state-mandated testing period, school lunches average 863 calories The economist claims that after the testing period ends, the average caloric content of Virginia school lunches drops significantly They collected a random sample of 500 students' school lunches around Virginia
a). What null and alternative hypotheses should you test?
b). Set up the rejection region for this study using alpha = 0.05 Interpret alpha = 0.05 in the words of the problem
c). Suppose the sample data yielded the test statistic z = -2.17 What conclusion can you draw for the test?
d). Calculate the observed p-value for the test statistic z = -2.17 Interpret the p-value and draw the conclusion based on it

Answer 1

Answer:

a) Null hypothesis: $\mu \geq 863$

Alternative hypothesis: $\mu >863$

b) For this case using the significance level of $\alpha=0.05$ we can use the normal standard distirbution in order to find a quantile who accumulates 0.05 of the area in the left and we got:

$z_{\alpha}=-1.64$

And the rejection zone would be:

$z<-1.64$

c) For this case since the statistic calculated is lower than the critical value we have enough evidence to reject the null hypothesis at 5% of significance

d) $p_v = P(z<-2.17) =0.015$

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis at the significance level provided

Step-by-step explanation:

Part a

We want to test for this case if the true mean is significantly less than 863 calories so then the system of hypothesis are:

Null hypothesis: $\mu \geq 863$

Alternative hypothesis: $\mu >863$

Part b

For this case using the significance level of $\alpha=0.05$ we can use the normal standard distirbution in order to find a quantile who accumulates 0.05 of the area in the left and we got:

$z_{\alpha}=-1.64$

And the rejection zone would be:

$z<-1.64$

Part c

For this case since the statistic calculated is lower than the critical value we have enough evidence to reject the null hypothesis at 5% of significance

Part d

For this case the p value would be given by:

$p_v = P(z<-2.17) =0.015$

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis at the significance level provided