Respuesta :
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 :-
[tex]t = \frac{x^{-} _{1} - x_{2} }{\sqrt{S^2(\frac{1}{n_{1} } }+\frac{1}{n_{2} } }[/tex]
where [tex]S^{2} = \frac{n_{1} S_{1} ^{2}+n_{2}S_{2} ^{2} }{n_{1} +n_{2} -2}[/tex]
[tex]S^{2} = \frac{20X(3.2)^2+24X(2.7)^2}{20+24-2}[/tex]
substitute values and we get S² = 40.988
[tex]t= \frac{17.53-14.89 }{\sqrt{40.988(\frac{1}{20} }+\frac{1}{24} )}[/tex]
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.
