Aconsumer 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, ?

We are confident at 99% that the difference between the two proportions is between

Step-by-step 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".  

represent the real population proportion for brand A  

represent the estimated proportion for Brand A

is the sample size required for Brand A

represent the real population proportion for brand b  

represent the estimated proportion for Brand B

is the sample size required for Brand B

represent the critical value for the margin of error  

The population proportion have the following distribution  

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

For the 99% confidence interval the value of and , with that value we can find the quantile required for the interval in the normal standard distribution.  

And replacing into the confidence interval formula we got:  

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

answer: no

step-by-step explanation: