On the morning of March 5, 1996, a train with 14 tankers of propane derailed nea

Question

On the morning of March 5, 1996, a train with 14 tankers of propane derailed near the center of the small Wisconsin town of Weyauwega, six of the tankers were ruptured and burning when the 1700 residents were ordered to evacuate the town. Researchers study this so that effective relief efforts can be designed for future disasters. About half of the households with pets did not evacuate their pets. A study conducted after the derailment focused on problems associated with retrieval of the pets after the evacuation and characteristics of the pet owners. One of the scales measured "commitment to adult animals, " and the people who evacuated all or some of their pets were compared with those who did not evacuate any of their pets. Higher scores indicate that the pet owner is more likely to take actions that benefit the pet. Here are the data summaries. Analyze the data and prepare a short report describing the results, (Use alpha = 0.01. Round your value for t to three decimal places and your P-value to four decimal places.) t = 13.1040 P-value = 3.4376 State your conclusion. Reject the null hypothesis. There is not significant evidence of a higher mean score for people who evaluated all or some pets. Fail to reject the null hypothesis. There is significant evidence of a higher mean score for people who evaluated all or some pets. Reject the null hypothesis. There is significant evidence of a higher mean score for people who evacuated all or some pets. Fail to reject the null hypothesis. There is not significant evidence of a higher mean score for people who evacuated all or some pets.

Explanation / Answer

The statistical software output for this problem is:

Two sample T hypothesis test:
1 : Mean of Population 1
2 : Mean of Population 2
1 - 2 : Difference between two means
H0 : 1 - 2 = 0
HA : 1 - 2 > 0
(with pooled variances)

Hypothesis test results:

Hence,

t = 3.459

p-value = 0.0003

Difference Sample Diff. Std. Err. DF T-Stat P-value 1 - 2 1.61 0.46540373 238 3.4593621 0.0003

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