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Ivanshal [37]
3 years ago
7

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

idents; it happens that the average salary of these 200 is about $40,000 with a SD of $12,000. Make a z -test of the null hypothesis that this difference was just chance (in the sampling).
Round your values to three decimal places.

The test statistic is ____ .
The P -value is approximately _____
Mathematics
1 answer:
Rina8888 [55]3 years ago
6 0

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.    

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