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

Wierzycaz 2020-10-28 Answered
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 who expressed interest in buying a tractor, 250 were randomly offered a warranty, and the rest did not receive. Ninety-three of those offering a warranty, and fifty-four of those not offering a warranty ended up buying 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 a manager offer a warranty based on this test?
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Expert Answer

Caren
Answered 2020-10-29 Author has 96 answers

Step 1
a)
All necessary assumptions are met:
The sampling method for each population is simple random sampling.
Independent samples.
Each sample includes at least 10 successes and 10 failures.
Proportion of customers purchasing tractors with warranties =p1=93250=0.372
Proportion of customers purchasing tractors without warranties =p2=54250=0.216
Difference in proportions =0.3720.216=0.156
Pooled sample proportion =p=(p1n1+p2n2)(n1+n2)=(0.372250+0.216250)500=0.294
The standard error (SE) of the sampling distribution difference between two proportions.
SE=p×(1p)×(1n1+1n2)
where p is the pooled sample proportion, n1 is the size of sample 1, and n2 is the size of sample 2.
SE={0.294(10.294)[(1/250)+(1/250)]}=0.04
Step 2
z value for 95% confidence interval is 1.96
Margin of error =zSE=1.960.04=0.0784
95% confidence interval for the difference between the proportions
=(0.1560.0784,0.156+0.0784)
=(0.077,0.235)
For 95% of the samples, the difference in proportions lies between 0.077 and 0.235
b)
Formulating Hypothesis
Null :H0: Offering the warranty does not increases the proportion of customers who eventually purchase a tractor.
Alternate : Ha: Offering the warranty increases the proportion of customers who eventually purchase a tractor.
Calculating Test statistic z :
Z=Difference in proportionsSE=0.1560.04
Z=3.9
Decision about the null hypothesis
At 0.05 significance level
Since it is observed that z∣=3.9>Zc=1.96, it is then concluded that the null hypothesis is rejected.
Using the P-value approach: The p-value is p=0.00004, and since P-vale <0.05, it is concluded that the null hypothesis is rejected.
Conclusion :
Offering the warranty increases the proportion of customers who eventually purchase a tractor.
Yes, they should offer a guarantee

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