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A random sample of 51 adult coyotes in a region of northern Minnesota showed the average age to be x = 2.03 years, with sample standard deviation s = 0.82 years. However, it is thought that the overall population mean age of coyotes is μ = 1.75. Do the sample data indicate that coyotes in this region of northern Minnesota tend to live longer than the average of 1.75 years? Use α = 0.01.

Respuesta :

Answer:

Yes the sample data   indicate that coyotes in this region of northern Minnesota tend to live longer than the average of 1.75 years

Step-by-step explanation:

From the question we are told that

   The sample size is  [tex]n = 51[/tex]

    The sample mean is  [tex]\= x = 2.03[/tex]

    The sample standard deviation is  [tex]\sigma = 0.82[/tex]

    The population mean is  [tex]\mu = 1.75[/tex]

    The level of significance is  [tex]\alpha = 0.01[/tex]

The null hypothesis  is  

      [tex]H_o : \mu = 0.82[/tex]

The alternative hypothesis is

     [tex]H_a : \mu >1.75[/tex]

The critical value of the the level significance [tex]\alpha[/tex] obtained from the critical value table for z-value is [tex]z_\alpha = 2.33[/tex]

 Now the test statistic is mathematically evaluated as

          [tex]t = \frac{\= x - \mu }{\frac{\sigma }{\sqrt{n} } }[/tex]

substituting values  

           [tex]t = \frac{ 2.03 - 1.75 }{\frac{0.82}{\sqrt{51} } }[/tex]

           [tex]t = 2.44[/tex]

From that calculated and obtained value we see that the critical value of the level of significance is less than the test statistics so we  reject the null hypothesis

Hence there sufficient evidence to proof that the sample data indicates that coyotes in this region of northern Minnesota tend to live longer than the average of 1.75 years