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 | 10 | 14 | 19 |
Afternoon shift | 10 | 9 | 14 | 16 |
At the .01 significance level, can we conclude there are more defects produced on the day shift? |
1. | State the decision rule. (Round your answer to 2 decimal places.) |
Reject H0 if t > |
2. | Compute the value of the test statistic. (Round your answer to 3 decimal places.) |
Value of the test statistic |
3. | What is the p-value? |
p-value | (Click to select) between 0.05 and 0.1 between 0.005 and 0.01 between 0.01 and 0.05 |
4. | What is your decision regarding H0? |
(Click to select) Do not reject Reject H0 |
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 12 12 16 19 Afternoon shift 10 10 12 15 At the .05 significance level, can we conclude there are more...
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 19 Afternoon shift 8 11 12 20 At the 0.050 significance level, can we conclude there are more defects produced on the day shift? Hint: For the...
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 14 18 Afternoon shift 9 10 13 16 At the .005 significance level, can we conclude there are more defects produced on the afternoon shift? Hint: For the...
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 10 16 17 Afternoon shift 9 10 14 15 At the 0.100 significance level, can we conclude there are more defects produced on the day shift? Hint: For the...
The null and alternate hypotheses are: HO: Haso H:Ha> 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 shift Afternoon shift 11 10 10 12 14 12 17 15 At the 0.100 significance level, can we conclude there are more defects produced on the day shit? Hint: For the calculations, assume the day shift as the first sample. a. State the...
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,...
The null and alternate hypotheses are: Ho HdsO 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 Day shift Afternoon shift 89 12 17 10 12 15 18 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. a. State the decision...
The null and alternative hypotheses are: H H :Hd < 0 :Hd > 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 Day shift Afternoon shift 1 10 8 2 12 9 3 15 12 4 11 15 At the 0.05 significance level, can we conclude there are more defects produced on the Afternoon shift? a. State the decision rule....
The null and alternate hypotheses are: H0 : μ1 = μ2 H1 : μ1 ≠ μ2 A random sample of 11 observations from Population 1 revealed a sample mean of 21 and sample deviation of 3.5. A random sample of 7 observations from Population 2 revealed a sample mean of 23 and sample standard deviation of 3.8. The underlying population standard deviations are unknown but are assumed to be equal. At the .05 significance level, is there a...
The null and alternate hypotheses are: H0 : μ1 = μ2 H1 : μ1 ≠ μ2 A random sample of 8 observations from Population 1 revealed a sample mean of 25 and sample deviation of 4.5. A random sample of 8 observations from Population 2 revealed a sample mean of 26 and sample standard deviation of 3.5. The underlying population standard deviations are unknown but are assumed to be equal. At the .05 significance level, is there a difference between...