Question

(CO6) In a sample of 15 stuffed animals, you find that they weigh an average of 8.56 ounces with a standard deviation of 0.09 ounces. Find the 92% confidence interval.

(8.526, 8.594)
(8.528, 8.591)
(8.510, 8.610)
(8.516, 8.604)

Answer 1

(8.510, 8.610)

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=8.56` represent the sample mean

`\mu` population mean (variable of interest)

s=0.09 represent the sample standard deviation

n=15 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 critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=15-1=14`

Since the Confidence is 0.98 or 98%, the value of
`\alpha=0.02` and
`\alpha/2 =0.01`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,14)".And we see that
`t_(\alpha/2)=2.62449`

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

`8.56-2.62449(0.09)/(√(15))=8.500`

`8.56-2.62449(0.09)/(√(15))=8.621`

So on this case the 95% confidence interval would be given by (8.500;8.621)

If we assume that the population deviation
`\sigma=0.09` then the confidence interval is given by:

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

Since the Confidence is 0.98 or 98%, the value of
`\alpha=0.02` and
`\alpha/2 =0.01`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.01,1,0)".And we see that
`z_(\alpha/2)=2.33`

Now we have everything in order to replace into formula (2):

`8.56-2.33(0.09)/(√(15))=8.510`

`8.56-2.33(0.09)/(√(15))=8.614`

So on this case the 95% confidence interval would be given by (8.510;8.610)

Based on this the most accurate answer would be:

(8.510, 8.610)