Can anybody help with this question PLZE!!

The average salary of all residents of a city is thought to be about $39,000. A research team surveys a random sample of 200 res

Rina8888 [55]

Answer:

$z=\frac{40000-39000}{\frac{12000}{\sqrt{200}}}=1.179$

$p_v =2*P(z>1.179)=2*[1-P(Z$

Step-by-step explanation:

Data given and notation

$\bar X=4000$ represent the average score for the sample

$s=12000$ represent the sample standard deviation

$n=200$ sample size

$\mu_o =39000$ represent the value that we want to test

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

z 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 apply a two tailed test.

What are H0 and Ha for this study?

Null hypothesis: $\mu = 39000$

Alternative hypothesis : $\mu \neq 39000$

Compute the test statistic

The sample size is large enough to assume the distribution for the statisitc normal. The statistic for this case is given by:

$z=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (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".

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

$z=\frac{40000-39000}{\frac{12000}{\sqrt{200}}}=1.179$

Give the appropriate conclusion for the test

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

$p_v =2*P(z>1.179)=2*[1-P(Z$

Conclusion

If we compare the p value and a significance level given for example $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, then the true population mean for the salary not differs significantly from the value of 39000.