Question

Is there a doctor in the house? A market research firm reported the mean annual earnings of all family practitioners in the United States was $178,258. A random sample of 50 family practitioners in Los Angeles had mean earnings of =x$192,920 with a standard deviation of $43,287. Do the data provide sufficient evidence to conclude that the mean salary for family practitioners in Los Angeles differs from the national average? Use the =α0.05 level of significance and the P-value method with the TI-84 calculator.

Answer 1

Answer:

$t=\frac{192920-178258}{\frac{43287}{\sqrt{50}}}=2.40$    

$p_v =2*P(t_{(49)}>2.40)=0.0202$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that th true mean is different from 178258 at 5% of signficance.  

TI 84 procedure

Press STAT and go to TESTS and select T-test

Enter the information and for the alternative use the not equal symbol

And then press on Calculate

Step-by-step explanation:

Data given and notation  

$\bar X=192920$ represent the sample mean

$s=43287$ represent the sample standard deviation

$n=50$ sample size  

$\mu_o =178258$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

$p_v$ 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 salary for family practitioners in Los Angeles differs from the national average , the system of hypothesis would be:  

Null hypothesis:$\mu = 178258$  

Alternative hypothesis:$\mu \neq 178258$  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

t-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".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

$t=\frac{192920-178258}{\frac{43287}{\sqrt{50}}}=2.40$    

P-value

The first step is calculate the degrees of freedom, on this case:  

$df=n-1=50-1=49$  

Since is a two side test the p value would be:  

$p_v =2*P(t_{(49)}>2.40)=0.0202$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that th true mean is different from 178258 at 5% of signficance.  

TI 84 procedure

Press STAT and go to TESTS and select T-test

Enter the information and for the alternative use the not equal symbol

And then press on Calculate