A random sample of 14 cows was selected from a large dairy herd at Brookfield Farm.

The milk yield in one week was recorded, in kg. The mean milk yield was 153.5 kg, with

a standard deviation of 23.5kg. Investigate the claim that the mean weekly milk yield for

the herd is greater than 120 kg. Evaluated on a TI 83 calculator

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Step 1It is given that the sample mean (\bar{x}) is 153.5 , standard deviation (s) is 23.5 and the sample size (n) is 14.Step 2The null and alternative hypotheses are shown below:\begin{array}{l}H_{0}: \mu=120 \\H_{\mathrm{a}}: \mu>120\end{array}Step-by-step procedure to find the test statistic and p-value using TI-83 calculator:Select STAT > Tests.Choose 2: T-Test...>StatsEnter \mu_{0} as 120, \bar{x} as 153.5, \mathbf{n} as 14.Select >\mu_{0}.Select Calculate and Click Enter.Output using TI- 83 calculator is shown below:From the output, it can be observed that the test statistic is 5.334 and the \mathrm{p}-value is 6.7856008^{*} 10^{-5}.Rejection rule:Reject the null hypothesis, if p-value is less than the level of significance.Conclusion:Consider the level of significance as 0.05 .Here, p-value is less than 0.05 .Thus, reject the null hypothesis.Hence, there is sufficient evidence to conclude that the mean weekly milk yield for the herd is greater than 120 \mathrm{~kg}. ...