A manager at ACME Equipment Sale and Rental wondered how offering a free two-yea

Question

A manager at ACME Equipment Sale and Rental wondered how offering a free two-year service warranty on its tractors might influence sales. For the next 500 customers expressing interest in purchasing a tractor, 250 were randomly offered the warranty and the rest were not. Ninety-three of those offered the warranty, and fifty-four of those not offered the warranty eventually purchased a tractor.

a. Construct a 95% confidence interval for the difference between the proportions of customers purchasing tractors with and without warranties. Be sure to check all necessary assumptions, and interpret the interval.

b. Test the hypothesis that offering the warranty increases the proportion of customers who eventually purchase a tractor. Be sure to check all necessary assumptions, state the null and alternative hypotheses, obtain the p-value, and state your conclusion. Should the manager offer the warranty based on this test?

c. Now suppose each tractor costs $15,000 and requires, on average, $7000 worth of work in the first two years. Is it worth it to the manager to offer the warranty? Hint: compare the expected profit for each potential tractor consumer offered the warranty with the expected profit for each potential tractor consumer not offered the warranty.

d. Revisit part (c). At what average repair cost in the first two years does it become profitable for the manager to offer the warranty?

Explanation / Answer

a.

All following conditions are met:

Proportions of customers purchasing tractors with warranties = 93 / 250 = 0.372

Proportions of customers purchasing tractors without warranties = 44 / 250 = 0.176

Difference in proportions = 0.372 - 0.176 = 0.196

Pooled sample proportion = p = (p1 * n1 + p2 * n2) / (n1 + n2) = (0.372 * 250 + 0.176 * 250) / 500 = 0.274

The standard error (SE) of the sampling distribution difference between two proportions.

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }

where p is the pooled sample proportion, n1 is the size of sample 1, and n2 is the size of sample 2.

SE = sqrt{ 0.274 * ( 1 - 0.274 ) * [ (1/250) + (1/250) ] } = 0.04

z value for 95% confidence interval is 1.96

Margin of error = z * SE = 1.96 * 0.04 = 0.0784

95% confidence interval for the difference between the proportions

= (0.196 - 0.0784, 0.196 + 0.0784)

= (0.1176, 0.2744)

For 95% of the samples, the difference in proportions lies between 0.1176 and 0.2744

b.

H0: Offering the warranty does not increases the proportion of customers who eventually purchase a tractor.

Ha: Offering the warranty increases the proportion of customers who eventually purchase a tractor.

Assumptions described in part(a) are satisfied.

Test statistic z = Difference in proportions / SE = 0.196 / 0.04 = 4.9

p-value = P(z > 4.9) is near 0.

As, p-value is less than the significance level of 0.05, we reject the null hypothesis and conclude that offering the warranty increases the proportion of customers who eventually purchase a tractor. So, the manager should offer the warranty based on this test.

c.

Proportions of customers purchasing tractors with warranties = 0.372

Proportions of customers purchasing tractors without warranties = 0.176

Expected profit with warranties = 0.372 * ($15,000 - $7000) = $2976

Expected profit without warranties = 0.176 * $15,000 = $2640

As, Expected profit with warranties is greater than Expected profit without warranties, it is worth to offer the warranty.

d.

Let C be the average repair cost in the first two year.

Expected profit with warranties = 0.372 * ($15,000 - C) > Expected profit without warranties = $2640

=> 5580 - 0.372 C > 2640

=> 0.372 C < 2940

=> C < 7903.226

So, at  average repair cost of $7903.23 in the first two years, it is profitable for the manager to offer the warranty.

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