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gogolik [260]
3 years ago
9

Which of the following is equivalent to the expression below?

Mathematics
1 answer:
vladimir1956 [14]3 years ago
5 0

Answer:

2i√5

Step-by-step explanation:

√-20 = √-1 × √20

√-1 = i

√20 = 2√5

i × 2√5 = 2i√5

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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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Answer: The probability is 0.5136.

To find this probability we have to find the range of the values on the normal distribution table.

First, we need to find the percent for each making z-score.

Lower end: 50/24 = 2.08 = 0.0183

Upper end: 2/24 = 0.08 = 0.5319

Now, that we have the upper and lower ends, we can subtract them to find the difference or the range.  

0.5319 - 0.0183 = 0.5136

There is a 51.36% chance they are in the given range.
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Answer:

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Step-by-step explanation:

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