Question

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.

Answer 1

a)
`100-1.68(25)/(√(50))=94.060`

`100+1.68(25)/(√(50))=105.940`

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

b)
`100-2.01(25)/(√(50))=92.894`

`100+2.01(25)/(√(50))=107.106`

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

c)
`100-2.68(25)/(√(50))=90.525`

`100+2.68(25)/(√(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)(s)/(√(n))` increase.

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)(s)/(√(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(25)/(√(50))=94.060`

`100+1.68(25)/(√(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(25)/(√(50))=92.894`

`100+2.01(25)/(√(50))=107.106`

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

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(25)/(√(50))=92.894`

`100+2.01(25)/(√(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(25)/(√(50))=90.525`

`100+2.68(25)/(√(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)(s)/(√(n))` increase.