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:
[tex]99\% \: \text {confidence interval} = (14.886, \: 16.113)\\\\[/tex]
Therefore, the 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is (14.886, 16.113) ounces.
Step-by-step 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
[tex]\bar{x} = 15.5[/tex]
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
[tex]$ s = 0.4957 $[/tex]
The confidence interval is given by
[tex]\text {confidence interval} = \bar{x} \pm MoE\\\\[/tex]
Where [tex]\bar{x}[/tex] is the sample mean and Margin of error is given by
[tex]$ MoE = t_{\alpha/2} \cdot (\frac{s}{\sqrt{n} } ) $ \\\\[/tex]
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
[tex]MoE = t_{\alpha/2}\cdot (\frac{s}{\sqrt{n} } ) \\\\MoE = 3.4994 \cdot \frac{0.4957}{\sqrt{8} } \\\\MoE = 3.4994\cdot 0.1753\\\\MoE = 0.6134\\\\[/tex]
So the required 99% confidence interval is
[tex]\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)\\\\[/tex]
Therefore, the 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is (14.886, 16.113) ounces.
