As the prices of heating oil and natural gas increase, consumers become more car

Question

As the prices of heating oil and natural gas increase, consumers become more careful

about heating their homes. Researchers want to know how warm homeowners keep their

houses in January and how the results from Wisconsin and Tennessee compare. The

researchers randomly call 23 Wisconsin households between 7 P.M. and 9 P.M. on January

15 and ask the respondent how warm the house is according to the thermostat. The

researchers then call 19 households in Tennessee the same night and ask the same

question. The results follow.

Wisconsin :

71 71 65 68

70 61 67 69

75 68 71 73

74 68 67 69

69 72 67 72

70 73 72

Tennessee:

73 75 74 71

74 73 74 70

72 71 69 72

74 73 70 72

69 70 67

For x=.01 , is the average temperature of a house in Tennessee significantly higher than

that of a house in Wisconsin on the evening of January 15? Assume the population

variances are equal and the house temperatures are normally distributed in each

population.

a) Decision is:

b) Test t =

c) p-value for this test is =

d) Critical t value(s) is/are

e) Copy and paste a little distribution diagram and label the critical t value(s)

f) Copy and paste a little distribution diagram and label the tail boundaries bounding the p-value

tails.

Explanation / Answer

Solution:-

a)

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

Null hypothesis: Wisconsin< Tennessee

Alternative hypothesis: Wisconsin > Tennessee

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the mean difference between sample means 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 = 0.8234

DF = 40

t = [ (x1 - x2) - d ] / SE

b) t = - 2.53

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 population means, and SE is the standard error.

We use the t Distribution Calculator to find P(t < - 2.53) = 0.0057

c) Therefore, the P-value in this analysis is 0.0057.

d) tcritical = 2.423

Interpret results. Since the P-value (0.0057) is less than the significance level (0.01), we have to reject the null hypothesis.

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