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

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:  

(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

Step-by-step explanation:

Data given and notation  

represent the mean average

represent the population standard deviation for the sample  

sample size  

represent the value that we want to test  

represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

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:  

Alternative hypothesis:  

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:  

(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

the expression (1.26)2t reveals the approximate rate of increases in the number of plants semiannually.

hope this : )