Question

In: Statistics and Probability

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

The null and alternate hypotheses are:

H0 : μd ≤ 0

H1 : μd > 0

The following sample information shows the number of defective units produced on the day shift and the afternoon shift for a sample of four days last month.

Day
1 2 3 4
Day shift 11 11 16 19
Afternoon shift 10 9 14 16

At the 0.010 significance level, can we conclude there are more defects produced on the day shift? Hint: For the calculations, assume the day shift as the first sample.

  1. State the decision rule. (Round your answer to 2 decimal places.)

  2. Compute the value of the test statistic. (Round your answer to 3 decimal places.)

  3. What is the p-value?

  • Between 0.005 and 0.01

  • Between 0.001 and 0.005

  • Between 0.01 and 0.025

  1. What is your decision regarding H0?

  • Reject H0

  • Do not reject H0

Solutions

Expert Solution

(a)

(b)

Test statistic = 4.899

(c)

Ans : Between 0.005 and 0.01

(d) Reject H0 (by decision rule)


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