Question

Question 5
Many stores sell extended warranties for products they sell. These are very lucrative for store owners. To learn more about who buys these warranties, a random sample was drawn of a store’s customers who recently purchased a product for which an extended warranty was available. Among other variables, each respondent reported whether he or she paid the regular price or a sale price and whether he or she purchased an extended warranty.

Regular Price Sale Price
Sample size 229 178
No who bought extended warranty 47 25
Can we conclude at the 10% significance level that those who paid the regular price are more likely to buy an extended warranty?

Answer 1

Answer on Question #44853 – Math - Statistics and Probability

Many stores sell extended warranties for products they sell. These are very lucrative for store owners. To learn more about who buys these warranties, a random sample was drawn of a store's customers who recently purchased a product for which an extended warranty was available. Among other variables, each respondent reported whether he or she paid the regular price or a sale price and whether he or she purchased an extended warranty.

Can we conclude at the 10% significance level that those who paid the regular price are more likely to buy an extended warranty?

Solution

$$H_0: p_1 - p_2 = 0 \quad H_a: p_1 - p_2 > 0$$ $$\widehat{p_1} = \frac{47}{229} = 0.205; \quad \widehat{p_2} = \frac{25}{178} = 0.140$$ $$\hat{p} = \frac{n_1 \widehat{p_1} + n_2 \widehat{p_2}}{n_1 + n_2} = \frac{229 \cdot 0.205 + 178 \cdot 0.140}{229 + 178} = 0.177$$ $$z = \frac{\widehat{p_1} - \widehat{p_2}}{\sqrt{\hat{p}(1 - \hat{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} = \frac{0.205 - 0.140}{\sqrt{0.177(1 - 0.177) \left(\frac{1}{229} + \frac{1}{178}\right)}} = 1.70.$$ $$p - value = P(Z > 1.70) = 1 - 0.9554 = 0.0446 < 0.05.$$

We reject $H_0$.

There is enough evidence to conclude that those who paid the regular price are more likely to buy an extended warranty.

www.AssignmentExpert.com