The null and alternate hypotheses are: H0 : μd ≤ 0 H1 : μd > 0...

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Statistics

The null and alternate hypotheses are: H0 : μd ≤ 0 H1 : μd >0 The following sample information shows the number of defectiveunits produced on the day shift and the afternoon shift for asample of four days last month. Day 1 2 3 4 Day shift 10 11 14 18Afternoon shift 8 11 12 15 At the 0.100 significance level, can weconclude there are more defects produced on the day shift? Hint:For the calculations, assume the day shift as the first sample.State the decision rule. (Round your answer to 2 decimal places.)Compute the value of the test statistic. (Round your answer to 3decimal places.) What is the p-value? Between 0.025 and 0.05Between 0.001 and 0.005 Between 0.005 and 0.01 What is yourdecision regarding H0? Reject H0 Do not reject H0

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The null and alternative hypothesis is given asthe true mean of difference in the number of defective unitsproduced in the Day shift and Afternoon shift is less than or equalto zerothe true mean of    See Answer
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