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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%.

User Batuhan
by
4.1k points

2 Answers

6 votes

Answer:


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

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


p_v =P(z<-1.307)=0.0956

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%.

Explanation:

Information provided

n=5000 represent the random sample taken

X=717 represent the surveys returned


\hat p=(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{(p_o (1-p_o))/(n)}} (1)

Replacing the info we got:


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

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


p_v =P(z<-1.307)=0.0956

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%.

User Danielassayag
by
4.5k points
4 votes

Answer:

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%.

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<0.15

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

Then, we can calculate the z-statistic as:


z=(p-\pi+0.5/n)/(\sigma_p)=(0.143-0.15+0.5/5000)/(0.005)=(-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<-1.366)=0.086

User Forellana
by
4.3k points