Question

Two sisters, sister A and sister B, play SCRABBLE with each other every evening. Sister A is a statistician, and she draws a random sample of 30 results from the 1,420 total games that have been played to construct a confidence interval estimate of p, the proportion of SCRABBLE games between her and her sister that she has won. Her 95% confidence interval estimate of p is LCL = 0.36, UCL = 0.69.

A 95% confidence interval estimate of the total number of games sister A has won out of the 1,450 games that have been played is LCL =___________ and UCL = ___________

The Illinois State Toll Highway Authority is conducting a study to estimate the proportion of low-income commuters who drive to work on a toll road. The project manager wants to estimate the proportion to within 0.03 with 95% confidence, and the project manager believes that p will turn out to be approximately 0.11.

A sample size no smaller than __________ is needed.

Answer 1

First question: LCL = 522, UCL = 1000.5

Second question: A sample size no smaller than 418 is needed.

Explanation:

First question:

Lower bound:

0.36 of 1450. So

0.36*1450 = 522

Upper bound:

0.69 of 1450. So

0.69*1450 = 1000.5

LCL = 522, UCL = 1000.5

Second question:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

The margin of error is:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

The project manager believes that p will turn out to be approximately 0.11.

This means that
`\pi = 0.11`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

The project manager wants to estimate the proportion to within 0.03

This means that the sample size needed is given by n, and n is found when M = 0.03. So

`M = z\sqrt{(\pi(1-\pi))/(n)}`

`0.03 = 1.96\sqrt{(0.11*0.89)/(n)}`

`0.03√(n) = 1.96√(0.11*0.89)`

`√(n) = (1.96√(0.11*0.89))/(0.03)`

`(√(n))^2 = ((1.96√(0.11*0.89))/(0.03))^2`

`n = 417.9`

Rounding up

A sample size no smaller than 418 is needed.