To test the hypothesis that the mean run time is more than 300 minutes, we can use a z-test. The null hypothesis (H0) is that the mean run time is equal to or less than 300 minutes, while the alternative hypothesis (Ha) is that the mean run time is greater than 300 minutes.
Given:
Sample size (n) = 50
Sample mean (x̄) = 305 minutes
Population standard deviation (σ) = 30 minutes
Significance level (α) = 0.05
First, we calculate the standard error of the mean (SE):
SE = σ/√n = 30/√50 ≈ 4.24
Next, we calculate the test statistic (z-score):
z = (x̄ - μ) / SE
μ is the hypothesized population mean, which is 300 minutes in this case.
z = (305 - 300) / 4.24 ≈ 1.18
To determine the critical value for a one-tailed test with a 0.05 significance level, we look up the z-value in the standard normal distribution table. The critical z-value is approximately 1.645.
Since the calculated z-value (1.18) is less than the critical z-value (1.645), we fail to reject the null hypothesis. There is not enough evidence to conclude that the inventor's claim of a mean run time of more than 300 minutes is true.
An inventor has developed a new, energy-efficient lawn mower
engine. He claims that the engine will run continuously for more
than 5 hours (300 minutes) on a single gallon of regular
gasoline. (The leading brand lawnmower engine runs for 300
minutes on 1 gallon of gasoline.)
From his stock of engines, the inventor selects a simple random
sample of 50 engines for testing. The engines run for an average
of 305 minutes. The true standard deviation σ is known and is
equal to 30 minutes, and the run times of the engines are
normally distributed.
Test hypothesis that the mean run time is more than 300
minutes. Use a 0.05 level of significance. By using z test
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