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love history [14]
3 years ago
6

A sample is chosen randomly from a population that was strongly skewed to the left. a) Describe the sampling distribution model

for the sample mean if the sample size is small. b) If we make the sample larger, what happens to the sampling distribution model’s shape, center, and spread? c) As we make the sample larger, what happens to the expected distribution of the data in the sample?
Mathematics
1 answer:
Nana76 [90]3 years ago
4 0

Answer:

a. The sampling distribution for the sample mean will be skewed to the left centered at the average u, and standard deviation will be  ∅/\sqrt{n}

b. The sample distribution will be normal in shape and will be centered at the average u, . standard deviation will be ∅/\sqrt{n}1

c. As the size of the sample increases, the sample distribution should draw near  and resemble the distribution of the population

Step-by-step explanation:

A sample is chosen randomly from a population that was strongly skewed to the left. a) Describe the sampling distribution model for the sample mean if the sample size is small. b) If we make the sample larger, what happens to the sampling distribution model’s shape, center, and spread? c) As we make the sample larger, what happens to the expected distribution of the data in the sample?

The following answers will march the questions above:

a. The sampling distribution for the sample mean will be skewed to the left centered at the average u, and standard deviation will be  ∅/\sqrt{n}

b. The sample distribution will be normal in shape and will be centered at the average u, . standard deviation will be ∅/\sqrt{n}1

c. As the size of the sample increases, the sample distribution should draw near  and resemble the distribution of the population

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A research firm needs to estimate within 3% the proportion of junior executives leaving large manufacturing companies within thr
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Answer:

972 junior executives should be surveyed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is of:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

35% of junior executives left their company within three years.

This means that \pi = 0.35

0.95 = 95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

To update this study, how many junior executives should be surveyed?

Within 3% of the proportion, which means that this is n for which M = 0.03. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.03 = 1.96\sqrt{\frac{0.35*0.65}{n}}

0.03\sqrt{n} = 1.96\sqrt{0.35*0.65}

\sqrt{n} = \frac{1.96\sqrt{0.35*0.65}}{0.03}

(\sqrt{n})^2 = (\frac{1.96\sqrt{0.35*0.65}}{0.03})^2

n = 971.2

Rounding up:

972 junior executives should be surveyed.

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3 years ago
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