The original price was $20, the sales price is $15. Is this an increase or decrease? What was the percent change?

6. Find the missing value. Show your work.

Rom4ik [11]

x = tan(13) * 24 =

0.2309 *24 = 5.5408

round answer as needed

5 0

3 years ago

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A recent study of two vendors of desktop personal computers reported that out of 836 units sold by Brand A, 111 required repair,

iris [78.8K]

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:

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