Question

Several years ago, a group of Honda Dealerships determined that the average amount of days automobiles stay on dealers' lots was equal to 22.1 days. They now believe that the mean time has changed and would like to test their belief at a significance level of 0.05. Results of a sample of 50 cars on various lots can be found in the file Honda Hypothesis Test. Use this information to answer the next several questions and ultimately decide whether or not the mean time has significantly changed.

data points

16 |

14 |

10 |

40 |

25 |

18 |

20 |

22 |

16 |

39 |

21 |

33 |

34 |

35 |

20 |

30 |

31 |

13 |

26 |

26 |

22 |

4 |

32 |

27 |

36 |

14 |

24 |

18 |

27 |

30 |

7 |

29 |

28 |

34 |

27 |

17 |

19 |

29 |

17 |

20 |

30 |

26 |

16 |

29 |

24 |

36 |

28 |

32 |

15 |

28 |

1. What should the alternative hypothesis be?

The mean of the sample is equal to 22.1 days.

The mean of the population is equal to 22.1 days.

The mean of the population is not equal to 22.1 days.

The mean of the sample is not equal to 22.1 days.

2.The correct critical value should be blank 1 ["t", "z"] = blank 2 ["1.86", "", "", "-1.68", "", "2.01", "1.86", "-1.64", "1.96"]

3. Find the test statistic, p-value, and critical value in terms of xbar.

4. conclusion

Since the test statistic is blank 1 ["greater than", "less than"] the positive critical value, blank 2 ["don't reject", "reject"] the null hypothesis. The number of days Hondas stay on the lots blank 3 ["has not changed.", "has changed."]

Answer #1

We are testing,

H0: u= 22.1 vs H1: u is not equal 22.1

1) so from above we can see that the alternative hypothesis is that the population mean is not equal to 22.1 days.

So option c) is correct.

2) Since Population sd is unknown, we will use t distribution here. we have n= 50, alpha= 0.05

So t critical value is at t alpha/2, n-1 ie at t 0.025,49=
**2.01**

3) From the sample dataset we have, sample mean= 24.48

Sample SD= 8.4594

Now test statistic: (24.48-22.1)/(8.4594/√50) = 1.989

p-value of this two sided t 49 test is:

2P(t49> 1.989) = 0.0523 (from t critical value tables)

Critical value found above: 2.01

4) Since the p-value of this test >significance level of 0.05, we have insufficient evidence to Reject H0.

We can also say that the test statistic <critical value, so we have insufficient evidence to Reject H0.

So we conclude that the number of days has not changed.

1. The following data show the number of days
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deviation and % coefficient of variation. Are there any
outliers (any values that exceed the mean by more than 3 standard
deviations)? If the expected value for the mean is 30
days, what is the percent bias and the sign? (12
points)
33 35 27 31 36
32 29 35 38 32
32 34 28 31 34
36 30 32 33 34

Dataset:
Husband Wife
53 50
38 34
46 44
30 36
31 23
26 31
29 25
48 51
65 46
29 26
28 23
29 32
25 22
22 21
21 24
43 50
24 27
38 33
33 26
35 29
40 29
27 25
27 24
49 38
48 39
27 27
26 24
40 34
35 37
21 26
32 39
30 28
60 52
36 40
25 23
66 55
52 34
41 40
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25
27
33
30
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25
33
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