Question

1. For the following problems:

Identify the claim and state the null and alternate hypotheses

Determine the critical value

Find the test statistic

Determine whether to reject or fail to reject the null hypothesis

Interpret the decision in the context of the original claim.

a. The store manager claims that the median number of customers per day through the checkout lines is no more than 225. A sample of customers per day through the checkout lines over 15 days is listed below. At .05 can you reject the manager’s claim?

215	224	261	208	194	198	230	216
213	200	154	223	210	174	187

b. A loan officer at a bank maintains that the median credit rating of its customers pursuing a mortgage loan is at least 695. The credit scores for 20 randomly selected mortgage loan customers is listed below. At .05 can you reject the loan officer’s claim?

617	695	706	631	711	625	653	612	707
719	605	619	621	699	644	665	697	609
687	711

c. A local police agency states that the median ticket cost for a speeding ticket issued is $185. In a random sample of 35 speeding tickets, the data is below. Can you reject the claim that the median ticket cost for a speeding ticket is $185?

154	158	135	157	185	100	178	140	177	97
111	99	115	156	147	102	140	175	185	114
142	128	224	159	131	187	167	145	120	218
195	201	130	194	221

Answer 1

SOLUTION a)

The provided number of positive signs is X+=2 and the number of negative signs is X−=13.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: Median = 225

Ha: Median less than 225

The test statistic is X=min{X+,X−}=min{2,13}=2.

Rejection Region

The critical value for the signficance level provided and the type of tail is left tailed so X∗=3.

Decision about the null hypothesis

Since in this case X=2≤X∗=3,

REJECT NULL HYPOTHESIS

There is enough evidence to claim that the population median number of customers per day through the checkout lines is no more than 225, at the 0.05 significance level.

b)

The provided number of positive signs is X+=7 and the number of negative signs is X−=12.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: Median = 695

Ha: Median > 695

The test statistic is X=min{X+,X−}=min{7,12}=7.

(2) Rejection Region

The critical value for the signficance level provided and the type of tail is X∗=5.

(3) Decision about the null hypothesis

Since in this case X=7>X∗=5,

DO NOT REJECT NULL HYPOTHESIS

There is not enough evidence to claim that the population median credit rating of its customers pursuing a mortgage loan is at least 695 at the 0.05 significance level.

c)

The provided number of positive signs is X^+ = 6X
+
=6 and the number of negative signs is X^- = 27X
−
=27.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: Median = $185

Ha: Median≠ $185

The test statistic is X=min{X+,X−}=min{6,27}=6.

Observe that the sample size n =35>25 is large enough to use normal approximation, so a z-statistic will be used.

Rejection Region: Based on the information provided, the significance level is \alpha = 0.05α=0.05, and the critical value for a two-tailed test is zc=1.96.

The rejection region for this two-tailed test is R={z:∣z∣>1.96}
(3) Test Statistics

The z-statistic is computed as follows:

Decision about the null hypothesis: Since it is observed that ∣z∣=3.482>zc=1.96, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0005, and since p=0.0005<0.05, it is concluded that the null hypothesis is rejected.

Conclusion: It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population median ticket cost for a speeding ticket is different than $185, at the α=0.05 significance level.