A national grocery store chain wants to test the difference in the average weight of turkeys sold in Detroit and the average weight of turke

Question

A national grocery store chain wants to test the difference in the average weight of turkeys sold in Detroit and the average weight of turkeys sold in Charlotte. According to the chain’s researcher, a random sample of 20 turkeys sold at the chain’s stores in Detroit yielded a sample mean of 17.53 pounds, with a sample standard deviation of 3.2 pounds. And a random sample of 24 turkeys sold at the chain’s stores in Charlotte yielded a sample mean of 14.89 pounds, with a sample standard deviation of 2.7 pounds. Use a 5% level of significance to determine whether there is a difference in the mean weight of turkeys sold in these two cities. Assume the population variances are approximately the same and use the pooled t-test

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Melody 3 months 2021-10-08T16:43:15+00:00 1 Answer 0 views 0

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    2021-10-08T16:44:29+00:00

    Answer:

    Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom

    The null hypothesis is accepted .

    Assume the population variances are approximately the same

    Step-by-step explanation:

    Explanation:-

    Given data a random sample of 20 turkeys sold at the chain’s stores in Detroit yielded a sample mean of 17.53 pounds, with a sample standard deviation of 3.2 pounds

    The first sample size  ‘n₁’= 20

    mean of the first sample ‘x₁⁻’= 17.53 pounds

    standard deviation of first sample  S₁ = 3.2 pounds

    Given data a random sample of 24 turkeys sold at the chain’s stores in Charlotte yielded a sample mean of 14.89 pounds, with a sample standard deviation of 2.7 pounds

    The second sample size  n₂ = 24

    mean of the second sample  “x₂⁻”= 14.89 pounds

    standard deviation of second sample  S₂ =  2.7 pounds

    Null hypothesis:-H₀: The Population Variance are approximately same

    Alternatively hypothesis: H₁:The Population Variance are approximately same

    Level of significance ∝ =0.05

    Degrees of freedom ν = n₁ +n₂ -2 =20+24-2 = 42

    Test statistic :-

       t = \frac{x^{-} _{1} -  x_{2} }{\sqrt{S^2(\frac{1}{n_{1} } }+\frac{1}{n_{2} }  }

       where         S^{2}   = \frac{n_{1} S_{1} ^{2}+n_{2}S_{2} ^{2}   }{n_{1} +n_{2} -2}

                          S^{2} = \frac{20X(3.2)^2+24X(2.7)^2}{20+24-2}

                 substitute values and we get  S² =  40.988

        t= \frac{17.53-14.89 }{\sqrt{40.988(\frac{1}{20} }+\frac{1}{24}  )}

        t =  1.3622

      Calculated value t = 1.3622

    Tabulated value ‘t’ =  2.081

    Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom

    Conclusion:-

    The null hypothesis is accepted

    Assume the population variances are approximately the same.

         

                           

                       

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