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.

sagnuhh 2020-10-26 Answered

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. StateA:162143150130152152143155131139164StateB:136140142131133129141140142136142

 
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Derrick
Answered 2020-10-27 Author has 94 answers

Null Hypothesis: H0: There is no sufficient evidence that the state A residents vary more than amounts from state B residents. Alternative Hypothesis: H0: 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: Ftest twoSample for VariancesStateAStateBMean147.5833333136.75Variance116.628787926.38636364Observations1212df1111F4.420040195p(Ff)onetail0.010391966F Critical onetail2.81793047 From the output, F = 4,42 P value = 0.10 Decision Rule: If P-value α, then reject the null hypothesis. Conclusion: Let consider the level of significance is α=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.

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