Question

A soft drink manufacturer wishes to know how many soft drinks adults drink each week. They want to construct a 95% confidence interval with an error of no more than 0.08. A consultant has informed them that a previous study found the mean to be 3.1 soft drinks per week and found the variance to be 0.49. What is the minimum sample size required to create the specified confidence interval? Round your answer up to the next integer.

Answer 1

The minimum sample size required to create the specified confidence interval is 295.

Explanation:

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

`\alpha = (1 - 0.95)/(2) = 0.025`

Now, we have to find z in the Z-table as such z has a p-value of
`1 - \alpha`.

That is z with a pvalue of
`1 - 0.025 = 0.975`, so Z = 1.96.

Now, find the margin of error M as such

`M = z(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

Variance of 0.49:

This means that
`\sigma = √(0.49) = 0.7`

They want to construct a 95% confidence interval with an error of no more than 0.08. What is the minimum sample size required to create the specified confidence interval?

The minimum sample size is n for which M = 0.08. So

`M = z(\sigma)/(√(n))`

`0.08 = 1.96(0.7)/(√(n))`

`0.08√(n) = 1.96*0.7`

`√(n) = (1.96*0.7)/(0.08)`

`(√(n))^2 = ((1.96*0.7)/(0.08))^2`

`n = 294.1`

Rounding up:

The minimum sample size required to create the specified confidence interval is 295.