Question

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 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

4. What is your decision regarding H0?

Answer #1

1) Reject H0 if t > 2.353

2)

value of the test statistic t=5.196

3) p-value =0.0069

4) reject Ho

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...

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
13
18
Afternoon shift
8
9
12
16
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:
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...

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 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
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...

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
9
10
14
15
At the 0.010 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 0.050 significance level, can we conclude there are more
defects produced on the day shift? Hint: For the...

The null and alternative hypotheses are:
H0:μd≤0H0:μd≤0
H1:μd>0H1: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
13
12
13
17
Afternoon shift
11
9
12
15
At the 0.10 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...

The null and alternative hypotheses are:
H0:μd≤0H0:μd≤0
H1:μd>0H1: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 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...

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
14
11
14
19
Afternoon shift
13
10
14
16
At the 0.01 significance level, can we conclude there are more
defects produced on the Afternoon shift?
b. Compute the value of the test statistic.
(Round the final...

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