Answer:
No the sample does not show that the average age  was different in 2010   Â
Step-by-step explanation:
From the question we are told that
  The sample size is n =  50
  The  sample mean is  [tex]\= x = 38.5[/tex]
  The population mean is  [tex]\mu = 37[/tex]
   The standard deviation is  [tex]\sigma = 16[/tex]
   The level of significance is  [tex]\alpha = 5 \% = 0.05[/tex]
 The null hypothesis is  [tex]H_o : \mu = 37[/tex]
 The alternative hypothesis is  [tex]H_a : \mu \ne 37[/tex]
The critical value of the level of  significance obtained from the normal distribution table is  ([tex]Z_{\alpha } = 1.645[/tex] )
Generally the test statistics is mathematically evaluated as
     [tex]t = \frac{ \= x - \mu }{ \frac{\sigma }{\sqrt{n} } }[/tex]
substituting values
     [tex]t = \frac{ 38.5 - 37}{ \frac{16}{\sqrt{50} } }[/tex]    Â
     [tex]t = 0.663[/tex]
Now looking at the value  t and  [tex]Z_{\alpha }[/tex] we see that [tex]t < Z_{\alpha }[/tex] 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 Â Â Â
