An article presents results of a survey of adults with diabetes. The average body mass index (BMI) in a sample of 1551 men was 30.4, with a standard deviation of 0.6. The average BMI in a sample of 1920 women was 31.1 with a standard deviation of 0.2. Let μX represent the population mean BMI in women with diabetes and let μY represent the population mean BMI in men with diabetes. Find a 99% confidence interval for the difference μX−μY. Round the answers to three decimal places.

So then the 99% confidence interval for is givenby (0.659; 0.741)

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".

We have the following data given:

represent the sample mean for women

the sample size of women

standard deviation for women

represent the sample mean for men

the sample size of  men

standard deviation for men

Solution to the problem

The confidence interval is given by the following formula:

   (1)

Since the sample sizes are large enough we can use the normal distribution as approximation of the t distribution. 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 formula (1) we got:

So then the 99% confidence interval for is givenby (0.659; 0.741)

the answer is two. hope this !