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 10 12 15 18 Afternoon shift 8 9 12 17

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.)

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

The table given below ,

 Day Day shift(X) Afternoon shift(Y) di=X-Y di^2 1 10 8 2 4 2 12 9 3 9 3 15 12 3 9 4 18 17 1 1 Sum 9 23

From table ,  The null and alternative hypothesis is ,  The test is right-tailed test.

Now , the df=degrees of freedom=n-1=4-1=3

The critical value is , ; From t-table

1. The decision rule is , Reject Ho if t-stat>3.18

2. The test statistic is , 3. The p-value is

between 0.005 to 0.01

Decision : Here , t-stat=4.700>3.18

Therefore , reject Ho.