Question

A recent study of two vendors of desktop personal computers reported that out of 836 units sold by Brand A, 111 required repair, while out of 739 units sold by Brand B, 111 required repair. Round all numeric answers to 4 decimal places.

Answer 1

Answer:

Step-by-step explanation:

Hello!

The study variables are:

$X_A$: The number of Brand A units sold that required repair.

$n_A= 836$

$x_A= 111$

$X_B$: THe number of Brand B units sold that required repair.

$n_B= 739$

$x_B= 111$

1. Calculate the difference in the sample proportion for the two brands of computers, p^BrandA−p^BrandB =?.

The sample proportion of each sample is equal to the number of "success" observed xi divided by the sample size n:

^p$_A$= $\frac{x_A}{n_A}= \frac{111}{836}= 0.1328$

^p$_B$= $\frac{x_B}{n_B}= \frac{111}{739} =0.1502$

^p$_A$ - ^p$_B$= 0.1328 - 0.1502= -0.0174

Note: proportions take numbers from 0 to 1, meaning they are always positive. But this time what you have to calculate is a difference between the two proportions so it is absolutely correct to reach a negative number it just means that one sample proportion is greater than the other.

2. What are the correct hypotheses for conducting a hypothesis test to determine whether the proportion of computers needing repairs is different for the two brands?

A. H0:pA−pB=0 , HA:pA−pB<0

B. H0:pA−pB=0 , HA:pA−pB>0

C. H0:pA−pB=0 , HA:pA−pB≠0

If you want to test whether the proportion of computers of both brands is different, you have to do a two-tailed test, the correct option is C.

3. Calculate the pooled estimate of the sample proportion, ^p= ?

To calculate the pooled sample proportion you have to use the following formula:

^p= $\frac{x_A+x_B}{n_A+n_B}= \frac{111+111}{836+739}= 0.14095 = 0.1410$

4. Is the success-failure condition met for this scenario?

A. Yes

B. No

The conditions that have to be met are:

$n_A\geq 30$ ⇒ Met

$n_A*p_A\geq 5$ ⇒ 836 * 0.1328= 111.4192; Met

$n_A*(1-p_A)\geq 5$ ⇒ 836 * (1 - 0.1328)= 727.5808; Met

$n_B\geq 30$ ⇒ Met

$n_B*p_B\geq 5$ ⇒ 739 * 0.1502= 110.9978; Met

$n_B*(1-p_B)\geq 5$ ⇒  739 * (1-0.1502)= 628.0022; Met

All conditions are met.

5. Calculate the test statistic for this hypothesis test. ? =

$Z_{H_0}= \frac{(p'_A-p'_B)-(p_A-p_B)}{\sqrt{p'(1-p')[\frac{1}{n_A} +\frac{1}{n_B} ]} } = \frac{-0.0174-0}{\sqrt{0.1410*0.859*[\frac{1}{836} +\frac{1}{739} ]} }= -0.9902$

6. Calculate the p-value for this hypothesis test, p-value = .

This hypothesis test is two-tailed and so is the p-value, since it has two tails you have to calculate it as:

P(Z≤-0.9902) + P(Z≥0.9902)=  P(Z≤-0.9902) + ( 1 - P(Z≤0.9902))= 0.161 + (1 - 0.839) = 0.322

7. What is your conclusion using α = 0.05?

A. Do not reject H0

B. Reject H0

The decision rule using th ep-value is:

If p-value > α, the decision is to not reject the null hypothesis.

If p-value ≤ α, the decision is to reject the null hypothesis.

The p-value= 0.322 is greater than α = 0.05, so the decision is to not reject the null hypothesis.

8. Compute a 95 % confidence interval for the difference p^BrandA−p^BrandB = ( , )

The formula to calculate the Confidence interval is a little different, because instead of the pooled sample proportion you have to use the sample proportion of each sample to calculate the standard deviation of the distribution:

($p'_A-p'_B$) ± $Z_{1-\alpha /2}$ * $\sqrt{\frac{p'_A(1-p'_A)}{n_A} +\frac{p'_B(1-p'_B)}{n_B} }$

-0.0174 ± 1.965 * $\sqrt{\frac{0.1328*0.8672}{836} +\frac{0.1502*0.8498}{739} }$

[-0.0520; 0.0172]

I hope it helps!