Question

A consumer advocate agency is concerned about reported failures of two brands of MP3 players, which we will label Brand A and Brand B. In a random sample of 197 Brand A players, 33 units failed within 1 year of purchase. Of the 290 Brand B players, 25 units were reported to have failed within the first year following purchase. The agency is interested in the difference between the population proportions, , for the two brands. What is the 99% confidence interval estimate of the true difference, ?

Answer 1

We are confident at 99% that the difference between the two proportions is between
`0.00056 \leq p_B -p_A \leq 0.16204`

Explanation:

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".

`p_A` represent the real population proportion for brand A

`\hat p_A =(33)/(197)=0.1675` represent the estimated proportion for Brand A

`n_A=197` is the sample size required for Brand A

`p_B` represent the real population proportion for brand b

`\hat p_B =(25)/(290)=0.0862` represent the estimated proportion for Brand B

`n_B=290` is the sample size required for Brand B

`z` represent the critical value for the margin of error

The population proportion have the following distribution

`p \sim N(p,\sqrt{(p(1-p))/(n)})`

The confidence interval for the difference of two proportions would be given by this formula

`(\hat p_A -\hat p_B) \pm z_(\alpha/2) \sqrt{(\hat p_A(1-\hat p_A))/(n_A) +(\hat p_B (1-\hat p_B))/(n_B)}`

For the 99% confidence interval the value of
`\alpha=1-0.99=0.01` and
`\alpha/2=0.005`, with that value we can find the quantile required for the interval in the normal standard distribution.

`z_(\alpha/2)=2.58`

And replacing into the confidence interval formula we got:

`(0.1675-0.0862) - 2.58 \sqrt{(0.0862(1-0.0862))/(290) +(0.1675(1-0.1675))/(197)}=0.00056`

`(0.1675 -0.0862) + 2.58 \sqrt{(0.0862(1-0.0862))/(290) +(0.1675(1-0.1675))/(197)}=0.16204`

And the 99% confidence interval would be given (0.00056;0.16204).

We are confident at 99% that the difference between the two proportions is between
`0.00056 \leq p_B -p_A \leq 0.16204`