Question

The null and alternate hypotheses are: |

H_{0} : μ ≤ 0_{d} |

H_{1} : μ > 0_{d} |

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 day shift? |

1. |
State the decision rule. (Round your answer to 2
decimal places.) |

Reject H_{0} 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 0.005 and 0.05 between 0.05 and 0.1 between 0.0005 and 0.005 |

4. |
What is your decision regarding H_{0}? |

(Click to select) Reject Do not
reject H_{0} |

Answer #1

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
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
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
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
12
12
16
19
Afternoon shift
10
10
12
15
At the .05 significance level, can we conclude there are more
defects produced...

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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