Question

Common information 29-33: An inventor has developed a new, energy-efficient lawn mower engine. He claims that the engine will run continuously for 5 hours (300 minutes) on a single gallon of regular gasoline. Suppose a simple random sample of 45 engines is tested. The engines run for an average of 290 minutes, assume the population standard deviation is 20 minutes. Test the null hypothesis that the mean run time is at least 300 minutes. Use a left sided test. Use a 0.025 level of significance. 29. Find the critical value for this test. 30. Find the test statistic 31. Find the p-value for this test. 32. a. My conclusion is Is the test statistically significant? b. 33. Could I use a symmetric confidence interval to solve this problem and if so what is it?

Answer 1

(29)

α = 0.025   

Critical z- score for a left-tailed test = -1.959963985   

(30)

Data:    

n = 45   

μ = 300   

s = 20   

x-bar = 290   

Hypotheses:    

Ho: μ ≥ 300   

Ha: μ < 300   

Test Statistic:    

SE = s/√n = 20/√45 = 2.98142397

z = (x-bar - μ)/SE = (290 - 300)/2.98142396999972 = -3.354101966

(31)

p- value = 0.000398115   

(32a and b)

Decision (in terms of the hypotheses):    

Since 0.000398115 < 0.025 we reject Ho

Conclusion (in terms of the problem):    

There is sufficient evidence that μ < 300

(33)

Since this is a left-tailed test, we can't use a symmetric (two-sided) confidence interval.

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