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:The standard error of the estimated difference is of 0.0313.Step-by-step explanation:To solve this question, we need to understand the central limit theorem, and subtraction of normal variables.Central Limit Theorem The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]. For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30. For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex] Subtraction between normal variables: When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances. Brand A:33 out of 197, so:[tex]p_A = \frac{33}{197} = 0.1675[/tex][tex]s_A = \sqrt{\frac{0.1675*0.8325}{197}} = 0.0266[/tex]Brand B:25 out of 290, so:[tex]p_B = \frac{25}{290} = 0.0862[/tex][tex]s_B = \sqrt{\frac{0.0862*0.9138}{290}} = 0.0165[/tex]What would be the standard error of the estimated difference?[tex]s = \sqrt{s_A^2+s_B^2} = \sqrt{0.0266^2+0.0165^2} = 0.0313[/tex]The standard error of the estimated difference is of 0.0313.