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A pharmaceutical manufacturer purchases a particular material from two different suppliers. The mean level of impurities in the raw material is approximately the same for

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A pharmaceutical manufacturer purchases a particular material from two different suppliers. The mean level of impurities in the raw material is approximately the same for both suppliers, but the manufacturer is concerned about the variability of the impurities from shipment to shipment. If the level of impurities tends to vary excessively for one source of supply, it could affect the quality of the pharmaceutical product. To compare the variation in percentage impurities for the two suppliers, the manufacturer selects 10 shipments from each of the two suppliers and measures the percentage of impurities in the raw material for each shipment. The sample means and variances are shown in the table. Supplier A Supplier B *1 = 1.88 *2 = 1.82 512 = 0.272 572 = 0.095 n1 = 10 n2 = 10 + (a) Do the data provide sufficient evidence to indicate a difference in the variability of the shipment impurity levels for the two suppliers? Test using a = 0.01. Based on the results of your test, what recommendation would you make to the pharmaceutical manufacturer? (Round your answers to two decimal places.) 1-2. Null and alternative hypotheses: O Ho: 012 + 022 versus Ha: 01 2 = 0,2 O Ho: 012 = 022 versus Ha: 012 > 022 O Ho: 012 02- O Ho: 0 2 = 072 versus Ha: 012 O Ho: 0 2 = 0,2 versus Ha: 012 5.35 X 5. Conclusion: O Ho is rejected. There is sufficient evidence to indicate that the supplier's shipments differ in variability. O Ho is not rejected. There is sufficient evidence to indicate that the supplier's shipments differ in variability. Ho is not rejected. There is insufficient evidence to indicate that the supplier's shipments differ in variability. O Ho is rejected. There is insufficient evidence to indicate that the supplier's shipments differ in variability. (b) Find a 99% confidence interval for 22. (Round your answers to three decimal places.) to Interpret your results. There is a 99% chance that an individual sample variation will fall within the interval

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