Mathematics, 07.04.2020 02:58 amandamelton1

A. A statistics practitioner took a random sample of 50 observations from a population with a standard deviation of 25 and computed the sample mean to be 100. Estimate the population mean with 90% confidence.
b. Repeat part (a) using a 95% confidence level.
c. Repeat part (a) using a 99% confidence level.
d. Describe the effect on the confidence interval estimate of increasing the confidence level.

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Answer from: ammerismccormick52

a) $100-1.68\frac{25}{\sqrt{50}}=94.060$

$100+1.68\frac{25}{\sqrt{50}}=105.940$

So on this case the 90% confidence interval would be given by (94.060;105.940)

b) $100-2.01\frac{25}{\sqrt{50}}=92.894$

$100+2.01\frac{25}{\sqrt{50}}=107.106$

So on this case the 95% confidence interval would be given by (92.894;107.106)

c) $100-2.68\frac{25}{\sqrt{50}}=90.525$

$100+2.68\frac{25}{\sqrt{50}}=109.475$

So on this case the 99% confidence interval would be given by (90.525;109.475)

d) When we increase the confidence level we see that the interval becomes wider and the margin of error given by $ME= t_{\alpha/2}\frac{s}{\sqrt{n}}$ increase.

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".

$\bar X=100$ represent the sample mean

$\mu$ population mean (variable of interest)

s=25 represent the sample standard deviation

n=50 represent the sample size

Part a

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

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

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

df=n-1=50-1=49

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 value. The excel command would be: "=-T.INV(0.05,49)".And we see that $t_{\alpha/2}=1.68$

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

$100-1.68\frac{25}{\sqrt{50}}=94.060$

$100+1.68\frac{25}{\sqrt{50}}=105.940$

So on this case the 90% confidence interval would be given by (94.060;105.940)

Part b

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,49)".And we see that $t_{\alpha/2}=2.01$

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

$100-2.01\frac{25}{\sqrt{50}}=92.894$

$100+2.01\frac{25}{\sqrt{50}}=107.106$

So on this case the 95% confidence interval would be given by (92.894;107.106)

Part b

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

$100-2.01\frac{25}{\sqrt{50}}=92.894$

$100+2.01\frac{25}{\sqrt{50}}=107.106$

So on this case the 95% confidence interval would be given by (92.894;107.106)

Part c

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,49)".And we see that $t_{\alpha/2}=2.68$

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

$100-2.68\frac{25}{\sqrt{50}}=90.525$

$100+2.68\frac{25}{\sqrt{50}}=109.475$

So on this case the 99% confidence interval would be given by (90.525;109.475)

Part d

When we increase the confidence level we see that the interval becomes wider and the margin of error given by $ME= t_{\alpha/2}\frac{s}{\sqrt{n}}$ increase.

Answer from: Quest

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17