if a smaple of 99 bowlers were taken from a population of 466 could refer to the mean of how many bowlers score
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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>
<em> where </em>
<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>
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