In 1998 the average income for middle class families in the US was $37,100 with a population standard deviation of $6362. We wanted to check how representative the 1998 GSS survey was of the American adult population from which it was drawn. The 1998 GSS survey found a mean income of $36,670 for its sample of 1225 middle-class American adults. Use this information to answer the following questions:______.
Which test statistic would be appropriate for this research problem?
A. Z test
B. T test
C. Chi square
D. None of the above

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

The best answer would be:

A. Z test

Step-by-step explanation:

Data given and notation

[tex]\bar X=36670[/tex] represent the mean average

[tex]\sigma=6362[/tex] represent the population standard deviation for the sample

[tex]n=1225[/tex] sample size

[tex]\mu_o =37100[/tex] represent the value that we want to test

[tex]\alpha[/tex] represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is equal or not to 37100, the system of hypothesis would be:

Null hypothesis:[tex]\mu =37100[/tex]

Alternative hypothesis:[tex]\mu \neq 37100[/tex]

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

The best answer would be:

A. Z test

codiepienagoya codiepienagoya · Answer 2 · 2021-11-22T17:06:04+01:00

Following are the calculation to the test statistics:

Given:

[tex]\mu = \$37,100 \\\\\bar{x}=\$36,670\\\\ \sigma= \$6362\\\\n=1225\\\\[/tex]

To find:

find the test statistic=?

Solution:

The population distribution is also missing in this research problem, as well as the standard deviation, which is similarly unclear.
This sample size is essentially normal whenever the sampling size is greater than 30.
Whenever someone wishes to use the z test to test the population mean, the community must be regular or the standard deviation should be known; otherwise, the Z test is used.
As just a result of not knowing the population variance, the suitable test statistic is indeed the "T-test".

Therefore, the final answer is "Option B".

Learn more:

brainly.com/question/16944173