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You wish to test the claim that mugreater than21 at a level of significance of alphaequals0.05 and are given sample statistics n equals 50 and x overbar equals 21.3. Assume the population standard deviation is 1.2. Compute the value of the standardized test statistic. Round your answer to two decimal places.

Respuesta :

Answer:

[tex]z = \frac{21.3-21}{\frac{1.2}{\sqrt{50}}}= 1.77[/tex]

Step-by-step explanation:

Data given and notation  

[tex]\bar X=21.3[/tex] represent the sample mean

[tex]\sigma=1.2[/tex] represent the population standard deviation

[tex]n=50[/tex] sample size represent the value that we want to test

[tex]\alpha[/tex] represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

[tex]p_v[/tex] 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 higher than 21, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq 21[/tex]  

Alternative hypothesis:[tex]\mu > 21[/tex]  

If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex]  (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

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

[tex]z = \frac{21.3-21}{\frac{1.2}{\sqrt{50}}}= 1.77[/tex]

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