# Listed below are amounts of strontium-90 (in millibecquerels or mBq per gram of calclum) in a simple random sample of baby teth obtained from resident of state A and state B.

Listed below are amounts of strontium-90 (in millibecquerels or mBq per gram of calclum) in a simple random sample of baby teth obtained from resident of state A and state B. Use a 0.05 significance level to test the celm that amounts of Strontium-90 from state A residents vary more than amounts from state B resints. Assume that both samples are independent simple random samples from populations having normal distributions. $\begin{array}{cccccccccccc}StateA:& 162& 143& 150& 130& 152& 152& 143& 155& 131& 139& 164\\ StateB:& 136& 140& 142& 131& 133& 129& 141& 140& 142& 136& 142\end{array}$

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Derrick

Null Hypothesis: ${H}_{0}:$ There is no sufficient evidence that the state A residents vary more than amounts from state B residents. Alternative Hypothesis: ${H}_{0}:$ There is a sufficient evidence that the state A residents vary more than amounts from state B residents. Test statistic: Excel Procedure: Enter the data for ‘State A’ and ‘State B’ in Excel sheet>Data>Data Analysis>F test two-samples for variances’ and click on ‘OK’>Select the column of ‘State A’ under ‘Variable 1 Range’>Select the column of ‘State B’ under ‘Variable 2 Range’>Click on ‘OK’. Exel Output: From the output, F = 4,42 P value = 0.10 Decision Rule: If P-value $\le \alpha$, then reject the null hypothesis. Conclusion: Let consider the level of significance is $\alpha =0.05$ Here, the p-value is less than the level of significance. From the rejection rule, reject the null hypothesis. Conclusion: There is a sufficient evidence that the state A residents vary more than amounts from state B residents.