A product testing laboratory selected a sample of 20 printers from two different manufacturers to determine if they could print more than 100,000 pages without breaking down. At $= .05$ , test the claim that the proportions of printers from Manufacturer 1 and Manufacturer 2 that lasted for more than $100,000 \text { pages are equal. }$
$\begin{array}{lcc} & \text { Manufacturer 1 } & \text { Manufacturer 2 } \\\text { Lasted past } 100,000 & 12 & 7 \\\text { Did not last to } 100,000 & 8 & 13 \\\text { Total } & 20 & 20\end{array}$

Question

A product testing laboratory selected a sample of 20 printers from two different manufacturers to determine if they could print more than 100,000 pages without breaking down. At $= .05$, test the claim that the proportions of printers from Manufacturer 1 and Manufacturer 2 that lasted for more than $100,000 \text { pages are equal. }$ $\begin{array}{lcc} & \text { Manufacturer 1 } & \text { Manufacturer 2 } \ \text { Lasted past } 100,000 & 12 & 7 \ \text { Did not last to } 100,000 & 8 & 13 \ \text { Total } & 20 & 20 \end{array}$ A) There is not evidence to reject the claim that the proportions are equal because the test value $2.506 <$ $3.841$ B) There is evidence to reject the claim that the proportions are equal because the test value $3.841 > 1.003$ C) There is not evidence to reject the claim that the proportions are equal because the test value $2.506 <$ $9.488$ D) There is not evidence to reject the claim that the proportions are equal because the test value $9.488 >$ $1.003$

Quizplus · Accepted Answer

To test the claim that the proportions of printers from Manufacturer 1 and Manufacturer 2 that lasted for more than 100,000 pages are equal, we will use a two-sample proportion test. The null hypothesis is that the proportions are equal, while the alternative hypothesis is that they are not equal.

Since the sample sizes are both greater than 10, we can use the normal approximation to the binomial distribution. The test statistic for the two-sample proportion test is calculated using the formula:

$$ z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}} $$

where $\hat{p}_1$ and $\hat{p}_2$ are the sample proportions for Manufacturers 1 and 2, $n_1$ and $n_2$ are the sample sizes, and $\hat{p}$ is the pooled sample proportion.

In this case,

$$ \hat{p}_1 = \frac{12}{20} = 0.6, \quad \hat{p}_2 = \frac{7}{20} = 0.35 $$

and

$$ \hat{p} = \frac{12+7}{20+20} = 0.425 $$

Plugging these values into the formula for the test statistic, we get:

$$ z = \frac{0.6 - 0.35}{\sqrt{0.425(1-0.425)\left(\frac{1}{20}+\frac{1}{20}\right)}} \approx 2.506 $$

To determine whether this test statistic is significant at the 0.05 level, we compare it to the critical value from the standard normal distribution. Since this is a two-tailed test, we will use a critical value of $\pm 1.96$.

Since $|2.506| < 1.96$, we do not reject the null hypothesis. This means that there is not enough evidence to conclude that the proportions of printers from Manufacturer 1 and Manufacturer 2 that lasted for more than 100,000 pages are different. Therefore, the best choice is A.

A Product Testing Laboratory Selected a Sample of 20 Printers =.05= .05=.05

A Product Testing Laboratory Selected a Sample of 20 Printers $= .05$