## 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

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

## Answers ( )

Answer:Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedomThe null hypothesis is accepted .Assume the population variances are approximately the sameStep-by-step explanation::-ExplanationGiven 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’= 20mean of the first sample ‘x₁⁻’= 17.53 poundsstandard deviation of first sample S₁ = 3.2 poundsGiven 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 = 24mean of the second sample “x₂⁻”= 14.89 poundsdsstandard deviation of second sample S₂ = 2.7 pounNull hypothesis:-H₀: The Population Variance are approximately sameAlternatively hypothesis: H₁:The Population Variance are approximately sameLevel of significance ∝ =0.05Degrees of freedom ν = n₁ +n₂ -2 =20+24-2 = 42Test statistic :-wheresubstitute values and we get S² = 40.988t = 1.3622Calculated 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

:-ConclusionThe null hypothesis is acceptedAssume the population variances are approximately the same.