In a simple random sample of 1200 Americans age 20 and over, there were 138 people with diabetes.
Report the 90% confidence interval for the proportion of all Americans age 20 and over with diabetes.

The 90% confidence interval for the proportion of all Americans age 20 and over with diabetes is between 0.0999 and 0.1301

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

For this problem, we have that:

90% confidence level

So , z is the value of Z that has a pvalue of , so .

The lower limit of this interval is:

The upper limit of this interval is:

The 90% confidence interval for the proportion of all Americans age 20 and over with diabetes is between 0.0999 and 0.1301

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it will cost dollar $540

the living room is 3 which means if we divide 325/3 then we get $108

therefore if 1part of the living room cost $108 and 3 costed $325

then total 5part of the room will cost 5*108 which gives $540