Help me please and

Graph: y = 12 - 41 + 2

tankabanditka [31]

Step-by-step explanation:

y=-27

...................

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

Earlier this month, 69% Californians voted 'yes' on Measure H: a sales tax measure to fund homeless services and prevention.Samp

VashaNatasha [74]

Answer:

0.59

Step-by-step explanation:

3 0

3 years ago

Julie says that she can perform a magic trick with numbers. She asks you to pick a whole number, any whole number. Square that n

Advocard [28]

Answer:

No

Step-by-step explanation:

I picked 2. Then got 4. Then 8. Then 9. Then 81. Then 79.(i dont know if this is right)

8 0

3 years ago

Why is the standard deviation usually preferred over the range?

konstantin123 [22]

Standard deviations, σ, is generally preferred over range. Reason being, it is calculated from all the data and when there is outliers it will not be impacted as much as the range. It is great to not also that, The IQR is preferred to the standard deviation when the distribution is very highly skewed or when there are severe outliers, because the IQR is less sensitive to this features than standard deviation.

4 0

3 years ago

Help me please and ​

Help me please and