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.

Group	n	x	s
Evacuated all or some pets	116	7.95	3.69
Did not evacuate any pets	125	6.26	3.55

Analyze the data and prepare a short report describing the results. (Use α = 0.01. Round your value for t to three decimal places and your P-value to four decimal places.)

t	=
P-value	=

State your conclusion.

Reject the null hypothesis. There is not significant evidence of a higher mean score for people who evacuated all or some pets.

Fail to 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.

Reject the null hypothesis. There is significant evidence of a higher mean score for people who evacuated all or some pets.

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u₁< u₂
Alternative hypothesis: u₁ > u₂

Note that these hypotheses constitute a one-tailed test.

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 = 0.46712
DF = 239

t = [ (x₁ - x₂) - d ] / SE

t = 3.62

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is thesize 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 population means, and SE is the standard error.

The observed difference in sample means produced a t statistic of 3.62

Therefore, the P-value in this analysis is 0.000.

Interpret results. Since the P-value (almost 0) is less than the significance level (0.01), hence we have to reject the null hypothesis.

Reject the null hypothesis. There is significant evidence of a higher mean score for people who evacuated all or some pets.