Question

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 is known and is equal to 30 minutes, and the run times of the engines are normally distributed. In R-studio, test the hypothesis that the mean run time is more than 300 minutes. Use a 0.01 level of significance.

Answer 1

Test statistic is 1.178

The mean run time is greater than 300 minutes

Explanation:

Null hypothesis: The mean run time is 300 minutes.

Alternate hypothesis: The mean run time is greater than 300 minutes.

Z = (sample mean - population mean) ÷ sd/√n = (305 - 300) ÷ 30/√50 = 5 ÷ 4.243 = 1.178

The test is a one-tailed test. Using a 0.01 significance level, the critical value is 2.326

Conclusion:

Reject the null hypothesis because the test statistic 1.178 is less than the critical value 2.326.

The mean run time is more than 300 minutes.

Answer 2

`z=(305-300)/((30)/(√(50)))=1.179`

`p_v =P(Z>1.179)=0.119`

If we compare the p value and the significance level given
`\alpha=0.01` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is higher than 300 at 1% of signficance.

Code at the end of the explanation.

Explanation:

Data given and notation

`\bar X=305` represent the sample mean

`\sigma=30` represent the population standard deviation for the sample

`n=50` sample size

`\mu_o =300` represent the value that we want to test

`\alpha=0.01` represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 300 min, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 300`

Alternative hypothesis:
`\mu > 300`

If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

`z=(\bar X-\mu_o)/((\sigma)/(√(n)))` (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

`z=(305-300)/((30)/(√(50)))=1.179`

P-value

Since is a one sided test the p value would be:

`p_v =P(Z>1.179)=0.119`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.01` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is higher than 300 at 1% of signficance.

R studio code

> library(BSDA)

> z.test(c(rep(305,50)),alternative = "greater",mu=300,sigma.x = 30)

One-sample z-Test

data: c(rep(305, 50))

z = 1.1785, p-value = 0.1193

alternative hypothesis: true mean is greater than 300

95 percent confidence interval:

298.0215 NA

sample estimates:

mean of x

305