Question

# 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 12 13 19 Afternoon shift 9 10 14 15

At the 0.025 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.)

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

1. What is the p-value?

• Between 0.05 and 0.10

• Between 0.001 and 0.005

• Between 0.005 and 0.01

1. What is your decision regarding H0?

• Do not reject H0

• Reject H0

Sol:

Rcode:

Dayshift <- c(11,12,13,19)
Afternoon <- c(9,10,14,15)
t.test(Dayshift,Afternoon,paired = TRUE,alternative = "greater")

Output:

Paired t-test

data: Dayshift and Afternoon
t = 1.6977, df = 3, p-value = 0.09406
alternative hypothesis: true difference in means is greater than 0
95 percent confidence interval:
-0.6757915 Inf
sample estimates:
mean of the differences
1.75

if p<0.025,Reject Ho

if p>0.025,Fail to reject Ho.

Compute the value of the test statistic. (

t=1.698

What is the p-value?

p value is  0.09406

Between 0.05 and 0.10

What is your decision regarding H0?

alpha=0.025

p=0.09406

p>alpha

Do not reject Ho.

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