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IceJOKER [234]
3 years ago
13

Suppose the weights of tight ends in a football league are normally distributed such that σ2=1,369. A sample of 49 tight ends wa

s randomly selected, and the weights are given below. Use Excel to calculate the 95% confidence interval for the mean weight of all tight ends in this league. Round your answers to two decimal places and use ascending order.
Mathematics
1 answer:
Darya [45]3 years ago
6 0

Answer:

Step-by-step explanation:Answer Explanation

Correct answers:

$\left(241.42,\ 262.14\right)$

A 95% confidence interval for μ is (x¯−zα/2σn‾√,x¯+zα/2σn‾√). Here, α=0.05, σ=37, and n=49. Use Excel to calculate the 95% confidence interval.

1. Open Excel, enter the given data in column A, and find the sample mean, x¯, using the AVERAGE function. Thus, the sample mean, rounded to two decimal places, is x¯=251.78.

2. Click on any empty cell, enter =CONFIDENCE.NORM(0.05,37,49), and press ENTER.

3. The margin of error, rounded to two decimal places, is zα/2σn‾√≈10.36. The confidence interval for the population mean has a lower limit of 251.78−10.36=241.42 and an upper limit of 251.78+10.36=262.14.

Thus, the 95% confidence interval for μ is (241.42, 262.14).

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A government report gives a 99% confidence interval for the proportion of welfare recipients who have been receiving welfare ben
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Answer:

The estimation for the proportion for this case is \hat p = 0.21

And we know that the 99% confidence interval is given by:

0.21 \pm 0.045

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}  

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Solution to the problem

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

The confidence interval would be given by this formula  

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}  

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=1.96

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=2.58

The estimation for the proportion for this case is \hat p = 0.21

And we know that the 99% confidence interval is given by:

0.21 \pm 0.045

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}  

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

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