A survey of 1000 air travelers1 found that prefer a window seat. The sample size is large enough to use the normal distribution, and a bootstrap distribution shows that the standard error is . Use a normal distribution to find a 90% confidence interval for the proportion of air travelers who prefer a window seat. Round your answers to three decimal places.

90% confidence interval for the true proportion of air travelers who prefer the window seat is (0.575, 0.625)

Step-by-step explanation:

We have the following data:

Sample size = n = 1000

Proportion of travelers who prefer window seat = p = 60%

Standard Error = SE = 0.015

We need to construct a 90% confidence interval for the proportion of travelers who prefer window seat. Therefore, we will use One-sample z test about population proportion for constructing the confidence interval. The formula to calculate the confidence interval is:

Since, standard error is calculated as:

Re-writing the formula of confidence interval:

Here, is the critical value for 90% confidence interval. From the z-table this value comes out to be 1.645.

Substituting all the values in the formula gives us:

Therefore, the 90% confidence interval for the true proportion of air travelers who prefer the window seat is (0.575, 0.625)

the camry sales+accord sale=828,000. to make it so the two numbers are 18 thousand apart and still equal 828,000, you first need to find out what half of 828,000 is. 828,000/2=414,000. for the two numbers to equal 828,000, you have to add the same amount as you subtract from the middle. in this case, the number is 9,000 because it's half of 18,000. to equal 828,000, you have to add 9,000 to 414,000 for the camry and subtract 9,000 from 414,000 for the accord. 414,000+9,000=423,000.

414,000-9,000=405,000.

423,000+405,000=828,000.

423,000-405,000=18,

hope it

3/5

step-by-step explanation:

double number line is not a standard concept, so it must have been defined in your textbook. assume top line is numerator.