A researcher wants to determine whether there is a relationship between age and mode of shopping. In a random sample of 18- to 21-year-olds, 140 of 185 people shopped online more often than in stores. In a random sample of 40- to 43-year-olds, 90 of 155 shopped online more often than in stores. What is the margin of error for a 95% confidence interval for the difference between the proportions of shoppers in these age groups who shop online more than in stores

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 18-21

represent the estimated proportion for 18-21

is the sample size required for 18-21

represent the real population proportion for 40-43

represent the estimated proportion for 40-43

is the sample size required for 40-43

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 the Margin of error is given by:

If we replace we got:

do you mean add both of the equations together

because if you do the answer is 10x-5

step-by-step explanation:

answer: okw

step-by-step explanation: