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, ?

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".

[tex]p_A[/tex] represent the real population proportion for brand A

[tex]\hat p_A =\frac{33}{197}=0.1675[/tex] represent the estimated proportion for Brand A

[tex]n_A=197[/tex] is the sample size required for Brand A

[tex]p_B[/tex] represent the real population proportion for brand b

[tex]\hat p_B =\frac{25}{290}=0.0862[/tex] represent the estimated proportion for Brand B

[tex]n_B=290[/tex] is the sample size required for Brand B

[tex]z[/tex] represent the critical value for the margin of error

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

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

[tex](\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}[/tex]

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

[tex]z_{\alpha/2}=2.58[/tex]

And replacing into the confidence interval formula we got:

[tex](0.1675-0.0862) - 2.58 \sqrt{\frac{0.0862(1-0.0862)}{290} +\frac{0.1675(1-0.1675)}{197}}=0.00056[/tex]

[tex](0.1675 -0.0862) + 2.58 \sqrt{\frac{0.0862(1-0.0862)}{290} +\frac{0.1675(1-0.1675)}{197}}=0.16204[/tex]

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 [tex]0.00056 \leq p_B -p_A \leq 0.16204[/tex]