Asurvey of 1000 students found that 274 chose professional baseball team a as their favorite team. in a similar survey involving 760 students, 240 of them chose team a as their favorite. compute a 95% con- fidence interval for the difference between the propor- tions of students favoring team a in the two surveys. is there a significant difference?

The 95% confidence interval would be given (-0.0851;0.00108).  

We are confident at 95% that the difference between the two proportions is between

Step-by-step explanation:

Previous concepts

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 team A 1

is the sample size

represent the real population proportion for team A 2  

represent the estimated proportion for team A 2

is the sample size required for team A 2

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 95% 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 95% confidence interval would be given (-0.0851;0.00108).  

We are confident at 95% that the difference between the two proportions is between

answer: sorry i can't you.

step-by-step explanation: