Question

A survey of 361 students recorded the number of hours of television they watched per week. The sample mean was 6.504 hours with a standard deviation of 5.584. The standard error of the mean was 0.294. Find a 90% confidence interval for the population mean.

Answer 1

The 90% confidence interval would be given by (6.022;6.986)

Explanation:

1) 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 =6.504` represent the sample mean for the sample

`\mu` population mean (variable of interest)

s=5.584 represent the sample standard deviation

n=361 represent the sample size

2) Confidence interval

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 mean and the sample deviation we can use the following formulas:

`\bar X= \sum_(i=1)^n (x_i)/(n)` (2)

`s=\sqrt{(\sum_(i=1)^n (x_i-\bar X))/(n-1)}` (3)

The mean calculated for this case is
`\bar X=6.504`

The sample deviation calculated
`s=5.584`

Since the sample size is large enough >30 we can use the z distirbution as approximation of the t distribution.

Since the Confidence is 0.90 or 90%, the value of
`\alpha=1-0.9=0.1` and
`\alpha/2 =0.05`, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that
`Z_(\alpha/2)=\pm 1.64`

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

`6.504-1.64(5.584)/(√(361))=6.022`

`6.504+1.64(5.584)/(√(361))=6.986`

So on this case the 90% confidence interval would be given by (6.022;6.986)