Question

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.

Answer 1

Answer:

$a)\ \ \bar x_m-\bar x_f=67.03\\\\b)\ \ E=15.7416\\\\c)\ \ CI=[51.2884, \ 82.7716]$

Step-by-step explanation:

a. -Given that:

$n_m=41\ , \ \sigma_m=33, \bar x_m=135.67\\\\n_f=37. \ \ ,\sigma_f=20, \ \ \bar x_f=68.64$

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

$\bar x_m-\bar x_f\\\\\therefore \bigtriangleup\bar x=135.67-68.64\\\\=67.03$

Hence, the pointer is estimator 67.03

b. The standard error of the point estimator,$\bar x_m-\bar x_f$ is calculated by the following following:

$\sigma_{\bar x_m-\bar x_f}=\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}$

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

$E=z_{\alpha/2}\times \sigma_{\bar x_m-\bar x_f}\\\\\\=z_{0.005}\times\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}\\\\\\=2.575\times \sqrt{\frac{33^2}{41}+\frac{20^2}{37}}\\\\\\=15.7416$

Hence, the margin of error is 15.7416

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

$\bar x_m-\bar x_f\ \pm z_{\alpha/2}\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}$

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

$CI=\bar x_m-\bar x_f\ \pm z_{\alpha/2}\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}\\\\=(135.67-68.64)\pm 15.7416\\\\=67.03\pm 15.7416\\\\=[51.2884, \ 82.7716]$

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

Answer 2

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

• The point estimate of the difference between the population mean expenditure for males and the population mean expenditure for females is of $67.03.

• The margin of error is of $15.74.

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

The standard errors are given by:

$s_M = \frac{33}{\sqrt{41}} = 5.1537$

$s_F = \frac{20}{\sqrt{37}} = 3.288$

The distribution of the difference is given by:

$\overline{x} = \mu_M - \mu_F = 135.67 - 68.64 = 67.03$

$s = \sqrt{s_M^2 + s_F^2} = \sqrt{5.1537^2 + 3.288^2} = 6.1132$

The margin of error of the interval is given by:

$M = zs$

We have to find the critical value, which is z with a p-value of $\frac{1 + \alpha}{2}$, in which $\alpha$ is the confidence level.

In this problem, $\alpha = 0.99$, thus, z with a p-value of $\frac{1 + 0.99}{2} = 0.995$, which means that it is z = 2.575.

Then, the margin of error is:

$M = zs = 2.575(6.1132) = 15.74$

The margin of error is of $15.74.

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

$\overline{x} - M = 67.03 - 15.74 = 51.29$

$\overline{x} + M = 67.03 + 15.74 = 82.77$

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

A similar problem is given at https://brainly.com/question/16184804