for the graph, complete the following (a) write a table of ordered pairs from the graph. (b) write an equation from the table. (
1 answer:
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Answer:
a)
The null hypothesis is .
The alternate hypothesis is .
The decision rule is: accept the null hypothesis for , reject the null hypothesis for
.
Since , we reject the null hypothesis and accept the alternate hypothesis that that percentage of athletes who graduate is less than for the student population at large.
b)
The p-value for this test is 0.0367. Since this p-value is less than the significance level of , we reject the null hypothesis and accept the alternate hypothesis that that percentage of athletes who graduate is less than for the student population at large.
Step-by-step explanation:
Question a:
Perform the appropriate hypothesis test to determine whether this is significant evidence that the percentage of athletes who graduate is less than for the student population at large:
At the null hypothesis, we test if the proportion is the same as the student population, of 91%. Thus:
At the alternate hypothesis, we test that the proportion for athletes is less than 91%, that is:
The test statistic is:
In which X is the sample mean, is the value tested at the null hypothesis,
is the standard deviation and n is the size of the sample.
Test if the proportion is less at the 0.05 level:
The critical value is z with a p-value of 0.05, that is, z = -1.645. Thus, the decision rule is: accept the null hypothesis for , reject the null hypothesis for
.
0.91 is tested at the null hypothesis:
This means that
A sports reporter contacted 152 athletes randomly sampled from that same university and time period and found that 132 of them had graduated within 6 years.
This means that
Value of the test statistic:
Since , we reject the null hypothesis and accept the alternate hypothesis that that percentage of athletes who graduate is less than for the student population at large.
(b) (3 points) Calculate the P-value for this test. Explain how this P-value can be use to test the hypotheses in part (a).
The p-value of the test is the probability of finding a sample proportion of 0.8684 or below. This is the p-value of z = -1.79.
Looking a the z-table, z = -1.79 has a p-value of 0.0367.
The p-value for this test is 0.0367. Since this p-value is less than the significance level of , we reject the null hypothesis and accept the alternate hypothesis that that percentage of athletes who graduate is less than for the student population at large.