Question

The amounts (in ounces) of juice in eight randomly selected juice bottles are: 15.3 15.3 15.7 15.7 15.3 15.9 15.3 15.9 Construct a 98% confidence interval for the mean amount of juice in all such bottles

Answer 1

`15.55-2.997(0.278)/(√(8))=15.26`

`15.55+2.997(0.278)/(√(8))=15.84`

The 98% confidence interval would be given by (15.26;15.84)

Explanation:

Notation

`\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 t_(\alpha/2)(s)/(√(n))` (1)

We can 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=15.55`

The sample deviation calculated
`s=0.278`

In order to calculate the critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=8-1=7`

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(0.01,7)".And we see that
`t_(\alpha/2)=2.997`

And the confidence interval is given by:

`15.55-2.997(0.278)/(√(8))=15.26`

`15.55+2.997(0.278)/(√(8))=15.84`

The 98% confidence interval would be given by (15.26;15.84)