The Edison Electric Institute has published figures on the annual number of kilo

Question

The Edison Electric Institute has published figures on the annual number of kilowatt-hours expended by various home appliances. It is claimed that a vacuum cleaner expends an average of 46 kilowatt-hours per year. If a random sample of 12 homes included in a planned study indicates that vacuum cleaners expend an average of 41 kilowatt-hours per year with a standard deviation of 10.9 kilowatt-hours, does this suggest at the 0.05 level of significance that vacuum cleaners expend, on the average, less than 46 kilowatt-hours annually? Assume the population of kilowatt-hours to be normal.

Explanation / Answer

Solution:-

The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

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

Null hypothesis: >= 46, vacuum cleaners expend, on the average, more than or equal to 46 kilowatt-hours annually
Alternative hypothesis: < 46, vacuum cleaners expend, on the average, less than 46 kilowatt-hours annually

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n) = 10.9 / sqrt(12) = 3.14655896708
DF = n - 1 = 12 - 1 = 11
t = (x - ) / SE = (41 - 46)/3.14655896708 = -1.589

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

Here is the logic of the analysis: Given the alternative hypothesis ( < 46), we want to know whether the observed sample mean is small enough to cause us to reject the null hypothesis.

The observed sample mean produced a t statistic test statistic of -1.589. We use the t Distribution Calculator to find P(t < -1.589) = 0.070184.

Interpret results. Since the P-value (0.070184) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Conclusion. We fail to reject the null hypothesis, we accept the null hypothesis, that is vacuum cleaners expend, on the average, more than or equal to 46 kilowatt-hours annually.

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