Question

Suppose the weights of tight ends in a football league are normally distributed such that σ2=400. A sample of 11 tight ends was randomly selected, and the weights are given below. Calculate the 95% confidence interval for the mean weight of all tight ends in this league. Round your answers to two decimal places and use ascending order.

Answer 1

`217.636-1.96(20)/(√(11))=205.817`

`217.636+1.96(20)/(√(11))=229.455`

So on this case the 95% confidence interval would be given by (205.82;229.46)

Explanation:

Assuming the following data: Weight 150 169 170 196 200 218 219 262 269 270 271

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)

`\sigma= √(400)= 20` represent the population standard deviation

n=11 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)

In order to calculate the mean we can use the following formula:

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

The mean calculated for this case is
`\bar X=217.636`

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: "=NORM.INV(0.025,0,1)".And we see that
`z_(\alpha/2)=1.96`

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

`217.636-1.96(20)/(√(11))=205.817`

`217.636+1.96(20)/(√(11))=229.455`

So on this case the 95% confidence interval would be given by (205.82;229.46)