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.

The 95% CI =  24.506 ≤ x≤ 24.314

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

Step-by-step explanation:

Here e have

μ = Population mean = 24.5 ounces

= Sample mean = 24.41 ounces

σ  = Standard deviation = 0.11 ounces

n = Sample size = 5 packages

At 95% confidence, we have

Where z = 1.96 and -z = -1.96

Therefore, the confidence interval is given by

CI =24.41 1.96 × =  24.506, 24.314 or

The 95% CI =  24.506 ≤ x≤ 24.314

When = μ, we have

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.

its a trick question

step-by-step explanation:

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