A family wants to rent a car to go on vacation. Company A charges $30.50 and 11¢ per mile. Company B charges $60.50 and 16¢ pe
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Answer:
<em>The calculated value |t| = 2.375 > 2.1009 at 0.05 level of significance</em>
<em>Null hypothesis is rejected </em>
<em>Alternative hypothesis is accepted</em>
<em>The extra carbonation of cola results in a higher average compression strength</em>
<u><em /></u>
Step-by-step explanation:
<u><em>Step(i):-</em></u>
<em>Given data</em>
<em>First sample size (n₁) = 10 </em>
<em>mean of the first sample(x₁⁻) = 537</em>
<em>standard deviation of the first sample (S₁) = 22</em>
<em>second sample size (n₂) = 10 </em>
<em>mean of the second sample (x₂⁻) = 559</em>
<em>standard deviation of the second sample (S₂) = 17</em>
<u><em>Step(ii)</em></u><em>:-</em>
<u><em>Null hypothesis : H₀:</em></u><em>- The extra carbonation of cola results in a lower average compression strength</em>
<u>Alternative Hypothesis :H₁</u>
<em>The extra carbonation of cola results in a higher average compression strength</em>
<u><em>Step(iii)</em></u><em>:-</em>
<em>By using student's t -test for difference of means</em>
<u><em>Test statistic</em></u>
<em> </em><em />
<em> where </em>
<em> </em><em />
<em> </em><em />
<em> </em><em />
<em> t = -2.375</em>
<em>|t| = |-2.375| = 2.375</em>
<em>Degrees of freedom</em>
<em>γ = n₁+n₂ -2 = 10+10-2 =18</em>
<em /><em />
<em>The calculated value |t| = 2.375 > 2.1009 at 0.05 level of significance</em>
<em>Null hypothesis is rejected </em>
<em>Alternative hypothesis is accepted</em>
<u><em>Final answer:-</em></u>
<em>The extra carbonation of cola results in a higher average compression strength</em>
<u><em /></u>
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<em> </em>
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