Question

a medical director found mean blood pressure x = 126.07 for an SRS of 72 executives. The standard deviation of the blood pressures of all executives is σ = 15. Give a 90% confidence interval for the mean blood pressure μ of all executives.

Answer 1

`126.07-1.64(15)/(√(72))=123.17`

`126.07+1.64(15)/(√(72))=128.97`

So on this case the 90% confidence interval would be given by (123.17;128.97)

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` represent the sample mean for the sample

`\mu` population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

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

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

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: "=-NORM.INV(0.05,0,1)".And we see that
`z_(\alpha/2)=1.64`

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

`126.07-1.64(15)/(√(72))=123.17`

`126.07+1.64(15)/(√(72))=128.97`

So on this case the 90% confidence interval would be given by (123.17;128.97)