ANSWER ASAP

A quarter of five less than a number is no more than a third of five more than the same number.

sertanlavr [38]

Answer:

I am pretty sure its a... im doing the same thing and i am kinda thinking its that.

Step-by-step explanation:

3 0

3 years ago

The National Collegiate Athletic Association (NCAA) requires colleges to report the graduation rates of their athletes. At one l

nevsk [136]

Answer:

a)

The null hypothesis is $H_0: p = 0.91$ .

The alternate hypothesis is $H_1: p < 0.91$ .

The decision rule is: accept the null hypothesis for $z > -1.645$ , reject the null hypothesis for $z < -1.645$ .

Since $z = -1.79 < -1.645$ , 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 $\alpha = 0.05$ , 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:

$H_0: p = 0.91$

At the alternate hypothesis, we test that the proportion for athletes is less than 91%, that is:

$H_1: p < 0.91$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ 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 $z > -1.645$ , reject the null hypothesis for $z < -1.645$ .

0.91 is tested at the null hypothesis:

This means that $\mu = 0.91, \sigma = \sqrt{0.91*0.09}$

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 $n = 152, X = \frac{132}{152} = 0.8684$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.8684 - 0.91}{\frac{\sqrt{0.91*0.09}}{\sqrt{152}}}$

$z = -1.79$

Since $z = -1.79 < -1.645$ , 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 $\alpha = 0.05$ , 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.

8 0

3 years ago

8 cleaners can clean an office in 3 hours,if the block needs to be cleaned in 2 hours,how many cleaners will be needed?

romanna [79]

Answer:

12 cleaners

Step-by-step explanation:

8 cleaner can clean in 3 hours = 1 office

1 cleaner can clean in 3 hours = $\frac{1}{8}$ part of job