Question

The null and alternative 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	9	9	13	15

At the 0.01 significance level, can we conclude there are more defects produced on the Afternoon shift?

a. State the decision rule. (Round the final answer to 3 decimal places.)

H₀ is click to select (rejected, not rejected) if t> _____

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

Test statistic ______

c. What is your decision regarding the null hypothesis?

H₀ should click to select (be rejected, not be rejected)

d. Determine the p-value. (Round the final answer to 4 decimal places.)

The p-value is _______

Answer 1

Answer:

Given that,

The null and alternative hypotheses are:

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.

S.No	Day
1	2	3	4
Day Shift	11	11	16	19
Afternoon Shift	9	9	13	15

At the 0.01 significance level.

Can we conclude there are more defects produced on the Afternoon shift:

At the 0.01 significance level, and df=n-1=4-1=3, the critical value is 4.541.

Decision rule:

Reject H_0 if the tets statistic is t > 4.541.

Calculation table:

S.No	Day(x1)	Afternoon(x2)	difference(d)=x1-x2	d^2
1	11	9	2	4
2	11	9	2	4
3	16	13	3	9
4	19	15	4	16
Total			d=11	d^2=33

Then,

The mean():

=d/n

=11/4

=2.75

The degrees of freedom (df)=n-1

=4-1

df=3

The standard deviation(s):

s=0.9575 (Approximately)

The standard error (SE):

=0.9575/2

SE=0.4788

The test statistic (t):

=(2.75-0)/0.4788

=2.75/0.4788

t=8.744(Approximately)

(a).

State the decision rule:

Decision rule:

Reject H_0 if the test statistic is t > 4.541.

(b).

Compute the value of the test statistic:

The value of the test statistic is t=8.744.

(c).

What is your decision regarding the null hypothesis:

Reject H_0.

(d).

Determine the p-value:

The p-value is 0.0016.

The result is significant at p < 0.01.