Question

The average age of a part-time seasonal employee at a Vail Resorts ski mountain has historically been 37 years. A random sample of 50 part-time seasonal employees in 2010 had a mean of 38.5 years with a standard deviation of 16 years. Required:a. At the 5 percent level of significance, does this sample show that the average age was different in 2010? b. Which is the right hypotheses to test the statement?c. What are the test statistic and critical value?

Answer 1

Answer:

No the sample does not show that the average age  was different in 2010      

Step-by-step explanation:

From the question we are told that

   The sample size is n =  50

    The  sample mean is  $\= x = 38.5$

    The population mean is  $\mu = 37$

     The standard deviation is  $\sigma = 16$

     The level of significance is  $\alpha = 5 \% = 0.05$

 The null hypothesis is  $H_o : \mu = 37$

  The alternative hypothesis is  $H_a : \mu \ne 37$

The critical value of the level of  significance obtained from the normal distribution table is  ($Z_{\alpha } = 1.645$ )

Generally the test statistics is mathematically evaluated as

          $t = \frac{ \= x - \mu }{ \frac{\sigma }{\sqrt{n} } }$

substituting values

          $t = \frac{ 38.5 - 37}{ \frac{16}{\sqrt{50} } }$        

         $t = 0.663$

Now looking at the value  t and  $Z_{\alpha }$ we see that $t < Z_{\alpha }$ hence we fail to reject the null hypothesis.

This mean that there is no sufficient evidence to state that the sample shows that the average age was different in 2010