Question

Use the given claim and test statistic to find the P - value. Claim: The mean monthly salary of employees at one company is at most $32,500. Test statistic: z = 0.52 0.6985 0.6030 0.3015 0.3970

Answer 1

Answer:

$p_v =P(z>0.52)=1-P(z<0.52)=1-0.6984=0.3015$  

Step-by-step explanation:

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean height actually is at most 32500, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 32500$  

Alternative hypothesis:$\mu > 32500$  

The reason is because the claim contains the equal sign since says "The mean monthly salary of employees at one company is at most $32,500" so it could be 32500 or less, and the alternative hypothesis never can have an equal sign.

The statistic is given by the following formula:

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

Calculate the statistic

The calculated value for the statistic is z =0.52, this value is given.

P-value

Since is a one right tailed test the p value would be:  

$p_v =P(z>0.52)=1-P(z<0.52)=1-0.6984=0.3015$