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777dan777 [17]
3 years ago
8

Computer Action Company sells computers and computer parts by mail. The company assures its customers that products are mailed a

s soon as possible after an order is placed with the company. The distribution of the times taken to mail products is known to follow a normal distribution. A sample of 28 recent orders showed that mean time taken to mail products for these orders was 70 hours and the standard deviation was 14 hours. If you were asked to find a 95% confidence interval for the mean time taken to mail products for all orders received at the office of this company, would you use z or t
Mathematics
1 answer:
Scorpion4ik [409]3 years ago
8 0

Answer:

The t distribution is used, because we have the standard deviation for the sample.

Step-by-step explanation:

When to use the z or the t distribution?

When we have the standard deviation for the population, we use the z distribution.

When we have the standard deviation for the sample, we use the t distribution.

In this question:

We have the standard deviation for the sample, which means that the t distribution is used.

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Softa [21]
74 is what percent of 95?


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UkoKoshka [18]

Answer:

a) The minimum sample size is 601.

b) The minimum sample size is 2401.

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:

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

For this problem, we have that:

We dont know the true proportion, so we use \pi = 0.5, which is when we are are going to need the largest sample size.

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.

a. If a 95% confidence interval with a margin of error of no more than 0.04 is desired, give a close estimate of the minimum sample size that will guarantee that the desired margin of error is achieved. (Remember to round up any result, if necessary.)

This is n for which M = 0.04. So

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

0.04 = 1.96\sqrt{\frac{0.5*0.5}{n}}

0.04\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.04}

(\sqrt{n})^2 = (\frac{1.96*0.5}{0.04})^2

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Rounding up

The minimum sample size is 601.

b. If a 95% confidence interval with a margin of error of no more than 0.02 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.

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

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(\sqrt{n})^2 = (\frac{1.96*0.5}{0.02})^2

n = 2401

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