Question

A simple random sample of 49 items resulted in a sample mean of 80. Assume a population standard deviation of 14.

a) Develop a 86.64% confidence interval about the mean.
b) Develop a 97.86% confidence interval about the mean.
c) What is the upper limit for a lower one-sided 86.43% confidence interval?
d) What is the lower limit for an upper one-sided 69.15% confidence interval?

Answer 1

a) The 86.64% confidence interval is given by (77.0;83.0)

b) The 97.86% confidence interval is given by (75.4;84.6)

c) The one side upper confidence interval is (80,82.2)

d) The one side lower confidence interval is (79,80)

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=80` represent the sample mean

`\mu` population mean (variable of interest)

`\sigma=14` represent the population standard deviation

n=49 represent the sample size

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

`\bar X \pm z_(\alpha/2)(\sigma)/(√(n))` (1)

a) Develop a 86.64% confidence interval about the mean.

Since the confidence is 0.8664 or 86.64%, the value of
`\alpha=1-0.8664=0.1336` and
`\alpha/2 =0.0668`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.0668,0,1)".And we see that
`z_(\alpha/2)=1.50`

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

`80-1.50(14)/(√(49))=77.0`

`80+1.50(14)/(√(49))=83.0`

So on this case the 86.64% confidence interval would be given by (77.0;83.0)

b) Develop a 97.86% confidence interval about the mean.

Since the confidence is 0.9786 or 97.86%, the value of
`\alpha=1-0.9786=0.0214` and
`\alpha/2 =0.0107`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.0107,0,1)".And we see that
`z_(\alpha/2)=2.30`

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

`80-2.30(14)/(√(49))=75.4`

`80+2.30(14)/(√(49))=84.6`

So on this case the 97.86% confidence interval would be given by (75.4;84.6)

c) What is the upper limit for a lower one-sided 86.43% confidence interval?

Since the confidence is 0.8643 or 86.43%, the value of
`\alpha=1-0.8643=0.1357`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.1357,0,1)".And we see that
`z_(\alpha/2)=1.0998`

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

`80+1.0998(14)/(√(49))=82.2`

So the one side upper confidence interval is (80,82.2)

d) What is the lower limit for an upper one-sided 69.15% confidence interval?

Since the confidence is 0.6915 or 69.15%, the value of
`\alpha=1-0.6915=0.3085`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=NORM.INV(0.3085,0,1)".And we see that
`z_(\alpha/2)=-0.5`

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

`80-0.5(14)/(√(49))=79`

So the one side lower confidence interval is (79,80)