Answer:
<em>Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom</em>
<em>The null hypothesis is accepted . </em>
<em>Assume the population variances are approximately the same</em>
<u><em>Step-by-step explanation:</em></u>
<u>Explanation</u>:-
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
<em>The first sample size 'n₁'= 20</em>
<em>mean of the first sample 'x₁⁻'= 17.53 pounds</em>
<em>standard deviation of first sample S₁ = 3.2 pounds</em>
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
<em>The second sample size n₂ = 24</em>
<em>mean of the second sample "x₂⁻"= 14.89 pounds</em>
<em>standard deviation of second sample S₂ = 2.7 poun</em>ds
<u><em>Null hypothesis</em></u><u>:-</u><u><em>H₀</em></u><em>: The Population Variance are approximately same</em>
<u><em>Alternatively hypothesis</em></u><em>: H₁:The Population Variance are approximately same</em>
<em>Level of significance ∝ =0.05</em>
<em>Degrees of freedom ν = n₁ +n₂ -2 =20+24-2 = 42</em>
<em>Test statistic :-</em>
<em> </em>![t = \frac{x^{-} _{1} - x_{2} }{\sqrt{S^2(\frac{1}{n_{1} } }+\frac{1}{n_{2} } }](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7Bx%5E%7B-%7D%20_%7B1%7D%20-%20%20x_%7B2%7D%20%7D%7B%5Csqrt%7BS%5E2%28%5Cfrac%7B1%7D%7Bn_%7B1%7D%20%7D%20%7D%2B%5Cfrac%7B1%7D%7Bn_%7B2%7D%20%7D%20%20%7D)
<em> where </em>![S^{2} = \frac{n_{1} S_{1} ^{2}+n_{2}S_{2} ^{2} }{n_{1} +n_{2} -2}](https://tex.z-dn.net/?f=S%5E%7B2%7D%20%20%20%3D%20%5Cfrac%7Bn_%7B1%7D%20S_%7B1%7D%20%5E%7B2%7D%2Bn_%7B2%7DS_%7B2%7D%20%5E%7B2%7D%20%20%20%7D%7Bn_%7B1%7D%20%2Bn_%7B2%7D%20-2%7D)
![S^{2} = \frac{20X(3.2)^2+24X(2.7)^2}{20+24-2}](https://tex.z-dn.net/?f=S%5E%7B2%7D%20%3D%20%5Cfrac%7B20X%283.2%29%5E2%2B24X%282.7%29%5E2%7D%7B20%2B24-2%7D)
<em> substitute values and we get S² = 40.988</em>
<em> </em>
<em></em>
<em> </em> 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
<u><em>Conclusion</em></u>:-
<em>The null hypothesis is accepted </em>
<em>Assume the population variances are approximately the same.</em>
<em> </em>
<em> </em>
<em> </em>