Question

The amounts (in ounces) of randomly selected eight 16-ounce beverage cans are given below. See Attached Excel for Data. Assume that the amount of beverage in a randomly selected 16-ounce beverage can has a normal distribution. Compute a 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans and fill in the blanks appropriately. A 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is ( , ) ounces. (round to 3 decimal places)

Answer 1

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

The amounts (in ounces) of randomly selected eight 16-ounce beverage cans are given below.

16.5, 15.2, 15.4, 15.1, 15.3, 15.4, 16, 15.1

Assume that the amount of beverage in a randomly selected 16-ounce beverage can has a normal distribution. Compute a 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans and fill in the blanks appropriately.

A 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is ( , ) ounces. (round to 3 decimal places)

Answer:

`99\% \: \text {confidence interval} = (14.886, \: 16.113)\\\\`

Therefore, the 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is (14.886, 16.113) ounces.

Explanation:

Let us find out the mean amount of the 16-ounce beverage cans from the given data.

Using Excel,

=AVERAGE(number1, number2,....)

The mean is found to be

`\bar{x} = 15.5`

Let us find out the standard deviation of the 16-ounce beverage cans from the given data.

Using Excel,

=STDEV(number1, number2,....)

The standard deviation is found to be

`$ s = 0.4957 $`

The confidence interval is given by

`\text {confidence interval} = \bar{x} \pm MoE\\\\`

Where
`\bar{x}` is the sample mean and Margin of error is given by

`$ MoE = t_(\alpha/2) \cdot ((s)/(√(n) ) ) $ \\\\`

Where n is the sample size, s is the sample standard deviation and is the t-score corresponding to a 99% confidence level.

The t-score corresponding to a 99% confidence level is

Significance level = α = 1 - 0.99 = 0.01/2 = 0.005

Degree of freedom = n - 1 = 8 - 1 = 7

From the t-table at α = 0.005 and DoF = 7

t-score = 3.4994

`MoE = t_(\alpha/2)\cdot ((s)/(√(n) ) ) \\\\MoE = 3.4994 \cdot (0.4957)/(√(8) ) \\\\MoE = 3.4994\cdot 0.1753\\\\MoE = 0.6134\\\\`

So the required 99% confidence interval is

`\text {confidence interval} = \bar{x} \pm MoE\\\\\text {confidence interval} = 15.5 \pm 0.6134\\\\\text {confidence interval} = 15.5 - 0.6134, \: 15.5 + 0.6134\\\\\text {confidence interval} = (14.886, \: 16.113)\\\\`

Therefore, the 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is (14.886, 16.113) ounces.