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Question

# Use either the critical-value approach or the P-value approach to perform the required hypothesis test. Response Insurance collects data on seat-belt use among U.S. drivers. Of 1000 drivers 25-34 years old, 27% said that they buckle up, whereas 330 of 1100 drivers 45-64 years old said that they did. At the 10% significance level, do the data suggest that there is a difference in seat-belt use between drivers 25-34 years old and those 45-64 years old? [SOURCE: USA TODAY Online]

Solution

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$H_0:p_1=p_2$

$H_a: p_1\neq p_2$

The sample proportion is the number of successes divided by the sample size:

$\hat{p}_1=\dfrac{x_1}{n_1}=27\%=0.27$

$\hat{p}_2=\dfrac{x_2}{n_2}=\dfrac{330}{1100}=0.30$

$\hat{p}_p=\dfrac{x_1+x_2}{n_1+n_2}=\dfrac{270+330}{1000+1100}=\dfrac{600}{2100}\approx 0.286$

Determine the value of the test statistic:

$z=\dfrac{\hat{p}_1-\hat{p}_2}{\sqrt{\hat{p}_p(1-\hat{p}_p)}\sqrt{\dfrac{1}{n_1}+\dfrac{1}{n_2}}}=\dfrac{0.27-0.30}{\sqrt{0.286(1-0.286)}\sqrt{\dfrac{1}{1000}+\dfrac{1}{1100}}}\approx -1.52$

Determine the P-value using table II:

$P=2\times 0.0643=0.1286$

If the P-value is smaller than the significance level, reject the null hypothesis:

$P>0.10=10\%\Rightarrow \text{ Do not reject } H_0$

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