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forsale [732]
3 years ago
15

Identify the property that is being used to solve the problem:

Mathematics
1 answer:
USPshnik [31]3 years ago
4 0

Answer:

Addition Property of equality

Step-by-step explanation:

The same number is added to both sides of the equations,

so the addition Property of equality is used.

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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,
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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!

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The LCM of 12 and 25 is 300
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