Wierzycaz

2020-10-28

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 $95\mathrm{%}$ 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?

Caren

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 $={p}_{1}=\frac{93}{250}=0.372$
Proportion of customers purchasing tractors without warranties $={p}_{2}=\frac{54}{250}=0.216$
Difference in proportions $=0.372-0.216=0.156$
Pooled sample proportion $=p=\frac{\left(p1\cdot n1+p2\cdot n2\right)}{\left(n1+n2\right)}=\frac{\left(0.372\cdot 250+0.216\cdot 250\right)}{500}=0.294$
The standard error (SE) of the sampling distribution difference between two proportions.
$SE=\sqrt{p×\left(1-p\right)×\left(\frac{1}{{n}_{1}}+\frac{1}{{n}_{2}}\right)}$
where p is the pooled sample proportion, ${n}_{1}$ is the size of sample 1, and ${n}_{2}$ is the size of sample 2.
$SE=\sqrt{\left\{0.294\cdot \left(1-0.294\right)\cdot \left[\left(1\text{/}250\right)+\left(1\text{/}250\right)\right]\right\}}=0.04$
Step 2
z value for $95\mathrm{%}$ confidence interval is 1.96
Margin of error $=z\cdot SE=1.96\cdot 0.04=0.0784$
$95\mathrm{%}$ confidence interval for the difference between the proportions
$=\left(0.156-0.0784,0.156+0.0784\right)$
$=\left(0.077,0.235\right)$
For $95\mathrm{%}$ of the samples, the difference in proportions lies between 0.077 and 0.235
b)
Formulating Hypothesis
Null :${H}_{0}$: 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 :
$Z=\frac{\text{Difference in proportions}}{SE}=\frac{0.156}{0.04}$
$Z=3.9$
At 0.05 significance level
Since it is observed that $\mid z\mid =3.9>{Z}_{c}=1.96$, 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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