Mathematics, 03.06.2020 05:57 khannie

In a study of cell phone use and brain hemispheric dominance, an Internet survey was e-mailed to 5000 subjects randomly selected from an online group involved with ears. 717 surveys were returned. Use a 0.01 significance level to test the claim that the return rate is less than 15%.

Answers: 3

Show answers

Answers

Answer from: gray4675

As the P-value (0.086) is greater than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the return rate is less than 15%.

Step-by-step explanation:

This is a hypothesis test for a proportion.

The claim is that the return rate is less than 15%.

Then, the null and alternative hypothesis are:

$H_0: \pi=0.15\\\\H_a:\pi$

The significance level is 0.01.

The sample has a size n=5000.

The sample proportion is p=0.143.

p=X/n=717/5000=0.143

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.15*0.85}{5000}}\\\\\\ \sigma_p=\sqrt{0.000026}=0.005$

Then, we can calculate the z-statistic as:

$z=\dfrac{p-\pi+0.5/n}{\sigma_p}=\dfrac{0.143-0.15+0.5/5000}{0.005}=\dfrac{-0.007}{0.005}=-1.366$

This test is a left-tailed test, so the P-value for this test is calculated as:

P-value=P(z

Answer from: aliciagore

$z=\frac{0.1434 -0.15}{\sqrt{\frac{0.15(1-0.15)}{5000}}}=-1.307$

Now we can calculate the p value with the following probability:

p_v =P(z

For this case the p value is lower than the significance level provided of 0.1 so then we can reject the null hypothesis and we can conclude that the the return rate is significantly less than 15%.

Step-by-step explanation:

Information provided

n=5000 represent the random sample taken

X=717 represent the surveys returned

$\hat p=\frac{717}{5000}=0.1434$ estimated proportion of urvyes returned

p_o=0.15 is the value to verify

$\alpha=0.01$ represent the significance level

z would represent the statistic

p_v represent the p value

System of hypothesis

We want to verify if the return rate is less than 15% the system of hypothesis are.:

Null hypothesis: $p \geq 0.15$

Alternative hypothesis: p < 0.15

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing the info we got:

$z=\frac{0.1434 -0.15}{\sqrt{\frac{0.15(1-0.15)}{5000}}}=-1.307$

Now we can calculate the p value with the following probability:

p_v =P(z

For this case the p value is lower than the significance level provided of 0.1 so then we can reject the null hypothesis and we can conclude that the the return rate is significantly less than 15%.

Answer from: Quest

answer: ouch. thats difficult

step-by-step explanation:

Answer from: Quest

a. -0.3 > -0.33

Another question on Mathematics

Mathematics, 21.06.2019 23:30

Express -5 2/5 as an improper fraction.

Answers: 2

Answer

Mathematics, 22.06.2019 00:00

If two parallel lines are cut by a transversal, interior angles on the same side of the transversal are supplementary. a. always b. sometimes c. never

Answers: 2

Answer

Mathematics, 22.06.2019 00:40

Point b lies between points a and c on . let x represent the length of segment ab in inches. use the segment to complete the statements. the value of x is . the length of in inches is . the length of in inches is .

Answers: 3

Answer

Mathematics, 22.06.2019 01:30

Aresearch study is done to find the average age of all u.s. factory workers. the researchers asked every factory worker in ohio what their birth year is. this research is an example of a census survey convenience sample simple random sample

Answers: 2

Answer

You know the right answer?

In a study of cell phone use and brain hemispheric dominance, an Internet survey was e-mailed to 500...

Questions