hw02_solution - 13.43 Define 1 as the average crop yield...

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13.43 Define 1 μ as the average crop yield using the current fertilizer. Define 2 as the average crop yield using the new fertilizer. Defile D as ( 2 - 1 ) (a) : 0 H 0 = D : 1 H 0 D Since the population variance (of the difference of the crop yield) is unknown, to test the population mean we use the t test here. We will check the required condition of using the t test later in part (c). Rejection condition of the test: For α = 0.05, we reject the null hypothesis if p-value < 0.05. Since the p-value = 0.1375 > 0.05, we fail to reject the null hypothesis. Thus we could conclude with 95% confidence that there is insufficient evidence to infer the new fertilizer is more effective than the current one. (b) We will check the required condition of using the t estimate later in part (c). The 95% confidence interval of the difference (in mean crop yields between the two fertilizers) is between -0.9157 and 2.9157. (c) To use the t test, the difference in mean crop yields has to be normal distributed. (d) To check whether the normality condition is satisfied, we draw the histogram of the sample difference.
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Histogram of the sample difference of the crop yields 5 1 2 4 0 0 2 4 6 -1.5 1 3.5 6 其他 Crop yield Frequency It seems that the histogram is bimodal. The required condition is not satisfied. (e) The data are experimental since the researchers influence and decide how the plots are treated. (f) If the land throughout the county was essentially the same, the researchers might randomly select some plots, treat half of them with the current fertilizer and treat the other with the new fertilizer, and finally record and test the difference in mean crop yields between the two fertilizers.
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13.47 Define 1 μ as the average annual sales among companies in this year. Define 2 as the average annual sales among companies in the last year. Defile as ( 1 - 2 ) (a) The 90% confidence of the improvement in sales between the two years is between 11.5890 and 27.9110 (in thousands of dollars) (b) : 0 H 0 = D : 1 H 0 D Since the population variance (of the difference of the sales) is unknown, to test the population mean we use the t test here. Rejection condition of the test: For α = 0.05, we reject the null hypothesis if p-value < 0.05. Since the p-value = 0.0001 < 0.05, we can reject the null hypothesis. Thus we could conclude with 95% confidence that there is sufficient evidence to infer companies advertise have higher sales than companies that do not. 20 60 100 其他 Sales Frequency
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hw02_solution - 13.43 Define 1 as the average crop yield...

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