Mathematics, 07.04.2020 22:47 jetblackcap

1. Assume the age of night school students is normally distributed. A simple random sample of 24 night school students had an average age of 27.3 years. Use this information and a sample standard deviation of 5.6 to find a a. 90% confidence interval estimate for the population variance b. 90% confidence interval estimate for the population standard deviation c. Explain the meaning of the confidence interval that you found in b. What does the 90% mean

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Answer from: sashamanger2396

a) $27.3-1.714\frac{5.6}{\sqrt{24}}=25.34$

$27.3+1.714\frac{5.6}{\sqrt{24}}=29.26$

So on this case the 90% confidence interval would be given by (25.34;29.26)

b) $\frac{(23)(5.6)^2}{35.172} \leq \sigma^2 \leq \frac{(23)(5.6)^2}{13.091}$

$20.507 \leq \sigma^2 \leq 55.097$

Now we just take square root on both sides of the interval and we got:

$4.528 \leq \sigma \leq 7.423$

So the best option would be:

4.528<σ<7.423

c) For this case the 90% means that we have 90% of confidence that the true parameter $\mu, \sigma$ is between the limits calculated on the intervals

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".

$\bar X=27.3$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=5.6 represent the sample standard deviation

n=24 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

Part a

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

df=n-1=24-1=23

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a talel to find the critical value. The excel command would be: "=-T.INV(0.05,23)".And we see that $t_{\alpha/2}=1.714$

Now we have everything in order to replace into formula (1):

$27.3-1.714\frac{5.6}{\sqrt{24}}=25.34$

$27.3+1.714\frac{5.6}{\sqrt{24}}=29.26$

So on this case the 90% confidence interval would be given by (25.34;29.26)

Part b

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,23)" "=CHISQ.INV(0.95,23)". so for this case the critical values are:

$\chi^2_{\alpha/2}=35.172$

$\chi^2_{1- \alpha/2}=13.091$

And replacing into the formula for the interval we got:

$\frac{(23)(5.6)^2}{35.172} \leq \sigma^2 \leq \frac{(23)(5.6)^2}{13.091}$

$20.507 \leq \sigma^2 \leq 55.097$

Now we just take square root on both sides of the interval and we got:

$4.528 \leq \sigma \leq 7.423$

So the best option would be:

4.528<σ<7.423

Part c

For this case the 90% means that we have 90% of confidence that the true parameter $\mu, \sigma$ is between the limits calculated on the intervals

Answer from: Quest

I’m pretty sure it’s i think : /

Answer from: Quest

6/ 11 = 0.545454 so your answer is55%

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