Question

Compute the value of the test statistic

The null and alternate hypotheses are: Ho Maso HP>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 2 12 10 1 10 9 Day shift Afternoon shift 13 14 4 18 15 At the 0.050 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. a. State the decision rule. (Round your answer to 2 decimal places.) Reject HO ift> 2.35 b. Compute the value of the test statistic. (Round your answer to 3 decimal places.) Value of the test statistic

Answer 1

Solution Gren let x = Day shift X2= Afternoon shift d = x,- Xz zreda X (d-Xd, 2 X2 d=x, - X2 To g 10 2 12 13 - 14 15 0.0625 0.5625 5.0625 3.0625 8.75 10 18 3. otal s Jet Ed=5 h=4 Selby the Ed n = 544 sulov-q Xd=1.25 20 Sd = v Eld-7d2 oli n-1 = V 8.75 3 Sa=1.7078 30

The null and alternative hypothesis is the Hoield co to Held> o X = 0.050 Decision Rulet tran-1) = t co.os 3) = 2.3534 from table. Reject Ho if t>2.35 (ъ) Test statistica (Xd-eld) t- 20 savn = (1-25-0) (1.7078) 25 t - 1.46387 t = 1.464) The value of the test statistic 1.464] pecision since To=1.464 < t (0.05,3) = 2.35 . Do not reject Ho.