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A marketing research company desires to know the mean consumption of milk per week among people over age 45. They believe that the milk consumption has a mean of 4.2 liters, and want to construct a 98% confidence interval with a maximum error of 0.08 liters. Assuming a standard deviation of 0.6 liters, what is the minimum number of people over age 45 they must include in their sample

Sagot :

Answer:

The minimum number of people over age 45 they must include in their sample is 305.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1 - 0.98}{2} = 0.01[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1 - \alpha[/tex].

That is z with a pvalue of [tex]1 - 0.01 = 0.99[/tex], so Z = 2.327.

Now, find the margin of error M as such

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

Standard deviation of 0.6 liters

This means that [tex]\sigma = 0.6[/tex]

What is the minimum number of people over age 45 they must include in their sample?

This is n for which M = 0.08. So

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

[tex]0.08 = 2.327\frac{0.6}{\sqrt{n}}[/tex]

[tex]0.08\sqrt{n} = 2.327*0.6[/tex]

[tex]\sqrt{n} = \frac{2.327*0.6}{0.08}[/tex]

[tex](\sqrt{n})^2 = (\frac{2.327*0.6}{0.08})^2[/tex]

[tex]n = 304.6[/tex]

Rounding up:

The minimum number of people over age 45 they must include in their sample is 305.