Question

Several years ago, a group of Honda Dealerships determined that the average amount of days automobiles...

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."]      

Homework Answers

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.

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