The USA Today reports that the average expenditure on Valentine's Day is $100.89. Do male and female consumers differ in the amounts they spend? The average expenditure in a sample survey of 41 male consumers was $135.67, and the average expenditure in a sample survey of 37 female consumers was $68.64. Based on past surveys, the standard deviation for male consumers is assumed to be $33, and the standard deviation for female consumers is assumed to be $20.
What is the point estimate of the difference between the population mean expenditure for males and the population mean expenditure for females (to 2 decimals)?
At 99% confidence, what is the margin of error (to 2 decimals)?
Develop a 99% confidence interval for the difference between the two population means (to 2 decimals). Use z-table.

Using the information given and the z-distribution, we have that:

The standard errors are given by:

The distribution of the difference is given by:

The margin of error of the interval is given by:

We have to find the critical value, which is z with a p-value of , in which is the confidence level.

In this problem, , thus, z with a p-value of , which means that it is z = 2.575.

Then, the margin of error is:

The margin of error is of $15.74.

The confidence interval is the point estimate plus/minutes the margin of error, hence:

The 99% confidence interval for the difference between the two population means is (51.29, 82.77).

A similar problem is given at link

Step-by-step explanation:

a. -Given that:

#The point estimator of the difference between the population mean expenditure for males and the population mean expenditure for females is calculated as:

Hence, the pointer is estimator 67.03

b. The standard error of the point estimator, is calculated by the following following:

-And the margin of error, E at a 99% confidence can be calculated as:

Hence, the margin of error is 15.7416

c. The estimator confidence interval is calculated using the following formula:

#We substitute to solve for the confidence interval using the standard deviation and sample size values in  a above:

Hence, the 99% confidence interval is [51.2884,82.7716]

well we know there are 4 clubs in 52 cards so it’s replaced so it should be 3/51

step-by-step explanation:

1/6

step-by-step explanation: