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telo118 [61]
1 year ago
15

A car insurance company suspects that the younger the driver is, the more reckless a driver he/she is. They take a survey and gr

oup their respondents based on age. Of 217 respondents between the ages of 16-25 (Group 1), 183 claimed to wear a seat belt at all times. Of 398 respondents who were 26+ years old (Group 2), 322 claimed to wear a seat belt at all times. Find a 95% confidence interval for the difference in proportions. Can it be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group?
Mathematics
1 answer:
erastovalidia [21]1 year ago
8 0

Answer:

The confidence interval for the difference in proportions is

-0.028\leq p_1-p_2 \leq 0.096

No. As the 95% CI include both negative and positive values, no proportion is significantly different from the other to conclude there is a difference between them.

Step-by-step explanation:

We have to construct a confidence interval for the difference of proportions.

The difference in the sample proportions is:

p_1-p_2=x_1/n_1-x_2/n_2=(183/217)-(322/398)=0.843-0.809\\\\p_1-p_2=0.034

The estimated standard error is:

\sigma_{p_1-p_2}=\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2} } \\\\\sigma_{p_1-p_2}=\sqrt{\frac{0.843*0.157}{217}+\frac{0.809*0.191}{398} } \\\\\sigma_{p_1-p_2}=\sqrt{0.000609912+0.000388239}=\sqrt{0.000998151} \\\\ \sigma_{p_1-p_2}=0.0316

The z-value for a 95% confidence interval is z=1.96.

Then, the lower and upper bounds are:

LL=(p_1-p_2)-z*\sigma_p=0.034-1.96*0.0316=0.034-0.062=-0.028\\\\\\UL=(p_1-p_2)+z*\sigma_p=0.034+1.96*0.0316=0.034+0.062=0.096

The confidence interval for the difference in proportions is

-0.028\leq p_1-p_2 \leq 0.096

<em>Can it be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group?</em>

No. It can not be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group, as the confidence interval include both positive and negative values.

This means that we are not confident that the actual difference of proportions is positive or negative. No proportion is significantly different from the other to conclude there is a difference.

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In the book Essentials of Marketing Research, William R. Dillon, Thomas J. Madden, and Neil H. Firtle discuss a research proposa
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Answer:

Null hypothesis:p_{1} = p_{2}  

Alternative hypothesis:p_{1} \neq p_{2}  

z=\frac{0.179-0.15}{\sqrt{0.17(1-0.17)(\frac{1}{140}+\frac{1}{60})}}=0.500  

p_v =2*P(Z>0.500)=0.617  

So the p value is a very low value and using any significance level for example \alpha=0.05, 0,1,0.15 always p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the two proportions NOT differs significantly.  

Step-by-step explanation:

Data given and notation  

X_{1}=25 represent the number of homeowners who would buy the security system

X_{2}=9 represent the number of renters who would buy the security system

n_{1}=140 sample 1

n_{2}=60 sample 2

p_{1}=\frac{25}{140}=0.179 represent the proportion of homeowners who would buy the security system

p_{2}=\frac{9}{60}= 0.15 represent the proportion of renters who would buy the security system

z would represent the statistic (variable of interest)  

p_v represent the value for the test (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the two proportions differs , the system of hypothesis would be:  

Null hypothesis:p_{1} = p_{2}  

Alternative hypothesis:p_{1} \neq p_{2}  

We need to apply a z test to compare proportions, and the statistic is given by:  

z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}   (1)  

Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{25+9}{140+60}=0.17  

Calculate the statistic  

Replacing in formula (1) the values obtained we got this:  

z=\frac{0.179-0.15}{\sqrt{0.17(1-0.17)(\frac{1}{140}+\frac{1}{60})}}=0.500  

Statistical decision

For this case we don't have a significance level provided \alpha, but we can calculate the p value for this test.    

Since is a two sided test the p value would be:  

p_v =2*P(Z>0.500)=0.617  

So the p value is a very low value and using any significance level for example \alpha=0.05, 0,1,0.15 always p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the two proportions NOT differs significantly.  

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