The null and alternate hypotheses are H0 d 0 H1 d 0 The

The null and alternate hypotheses are:

  

H₀ : ?_d ? 0

H₁ : ?_d > 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

10    

12    

13   

18   

  Afternoon shift

8    

9    

12   

16   

At the .10 significance level, can we conclude there are more defects produced on the afternoon 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 H₀ if t >

(b)

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

  Value of the test statistic

(c)

What is the p-value?

  p-value

(d)

What is your decision regarding H₀?

  

H₀

The null and alternate hypotheses are:

H0 : ?d ? 0

H1 : ?d > 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.

Solution

(a) The degree of freedom =n-1=4-1=3

Given a=0.1, the critical value is t(0.1, df=3) =1.64 (from student t table)

Reject H₀ if t >1.64

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(b)

the value of the test statistic

t=mean difference/(s/vn)

=4.899

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(c) the p-value is 0.0081 (from student t table)

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(d) Reject Ho.

13.250	mean Day shift
11.250	mean Afternoon Shift
2.000	mean difference (Day shift - Afternoon Shift)
0.816	std. dev.
0.408	std. error
4	n
3	df