Question

Your company prepares and distributes frozen foods. The package claims a net weight of 24.5 ounces. A random sample of today's production was weighted, and the results were summarized as follows: average=24.41 ounces; standard deviation=0.11 ounce. Sample size=5 packages. Find the 95% confidence interval for the mean weight you would have found had you weighed all packages produced today. Interpret your results.

Answer 1

The 95% CI = 24.506 ≤ x≤ 24.314

The 95% CI is 0.9 less than the package claim at CI
`_(claim)` = 24.404 ≤ x≤ 24.596

Explanation:

Here e have

μ = Population mean = 24.5 ounces

`\bar {x}` = Sample mean = 24.41 ounces

σ = Standard deviation = 0.11 ounces

n = Sample size = 5 packages

At 95% confidence, we have

`CI=\bar{x}\pm z(\sigma)/(√(n))`

Where z
`_c` = 1.96 and -z
`_c` = -1.96

Therefore, the confidence interval is given by

CI =24.41
`\pm` 1.96 ×
`(0.11)/(√(5))` = 24.506, 24.314 or

The 95% CI = 24.506 ≤ x≤ 24.314

When
`\bar {x}` = μ, we have

`CI=\mu \pm z(\sigma)/(√(n))` which gives

CI = 24.404 ≤ x≤ 24.596

This shows that at 95 % confidence level, the observed mean of the packages produced on the day is approximately equivalent to the package claim or 0.09 less than package claim.

Answer 2

`24.41-2.776(0.11)/(√(5))=24.27`

`24.41+ 2.776(0.11)/(√(5))=24.55`

So on this case the 95% confidence interval would be given by (24.27;24.55)

And for this case since the value if 24.5 is included in the confidence interval we don't have enough evidence to conclude that the true mean is significantly higher than 24.5 ounces

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=24.41` represent the sample mean

`\mu` population mean (variable of interest)

s=0.11 represent the sample standard deviation

n=5 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=5-1=4`

Since the Confidence is 0.95 or 95%, the value of
`\alpha=0.05` and
`\alpha/2 =0.025`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,4)".And we see that
`t_(\alpha/2)=2.776`

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

`24.41-2.776(0.11)/(√(5))=24.27`

`24.41+ 2.776(0.11)/(√(5))=24.55`

So on this case the 95% confidence interval would be given by (24.27;24.55)

And for this case since the value if 24.5 is included in the confidence interval we don't have enough evidence to conclude that the true mean is significantly higher than 24.5 ounces