Question

A sample of 16 elements is selected from a population with a reasonably symmetrical distribution. The sample mean is 100 and the sample standard deviation is 40. We can say that we are 90% confident that is between what two numbers? 12. What is the upper number? Maintain three decimal points in your calculations and give at least two decimal points in your answer.

Answer 1

`100-1.753(40)/(√(16))=82.470`

`100+ 1.753(40)/(√(16))=117.530`

So on this case the 90% confidence interval would be given by (82.470;117.530)

And the upper value would be 117.530 for this case

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=40 represent the sample standard deviation

n=16 represent the sample size

Solution to the problem

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=16-1=15`

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(0.05,15)".And we see that
`t_(\alpha/2)=1.753`

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

`100-1.753(40)/(√(16))=82.470`

`100+ 1.753(40)/(√(16))=117.530`

So on this case the 90% confidence interval would be given by (82.470;117.530)

Answer 2

We can say that we are 90% confident that the population mean is between 29.876 and 170.124.

The upper number is 170.124.

Explanation:

We have the sample's standard deviation, so we use the students' t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 16 - 1 = 15

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 15 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.9)/(2) = 0.95([tex]t_(95)`). So we have T = 1.7531

The margin of error is:

M = T*s = 40*1.7531 =70.124

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 100 - 70.124 = 29.876

The upper end of the interval is the sample mean added to M. So it is 100 + 70.124 = 170.124

We can say that we are 90% confident that the population mean is between 29.876 and 170.124.

The upper number is 170.124.