On the morning of March 5, 1996, a train with 14 tankers of propane derailed nea
ID: 3206415 • Letter: O
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 disasters like this so that effective relief efforts can be designed for future disasters. About half of the households with pets did not evacuate all of 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 = 0.01. Round your value for t to three decimal places and your P-value to four decimal places.)
Group n x s Evacuated all or some pets 112 7.84 3.68 Did not evacuate any pets 125 6.33 3.55Explanation / Answer
Solution:-
x1 = 7.84, s1 = 3.68, n1 = 112
x2 = 6.33, s2 = 3.55, n2 = 125
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1 - 2 = 0
Alternative hypothesis: 1 - 2 0
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.01. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 0.471
DF = 230
t = [ (x1 - x2) - d ] / SE
t = 3.2059
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.
Thus, the P-value = 0.0015
Interpret results. Since the P-value (0.001537) is less than the significance level (0.10), we cannot accept the null hypothesis.
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