Question

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 simple random samples of 500 engines were tested. The engines run for an average of 295 minutes, with a standard deviation of 20 minutes. Test the claim that the run time differs from 300 minutes. Use a 0.05 level of significance a) Set up test hypothesis b) Write test statistics and sketch c) Apply classical or P-value method d) Draw conclusion and interpret.

Answer 1

Answer:

a) The null and alternative hypothesis are:

$H_0: \mu\geq 300\\\\H_1: \mu$

b) test statistic t=-5.59 (df=499).

c) P-value=0

d) Reject the null hypothesis. The sample is enough evidence to reject the inventor's claim and to conclude that the run time is lower than 300 minutes.

Step-by-step explanation:

We have to perform an hypothesis test on the mean, with estimated standard deviation. We want to test the claim that the run time differs from 300 minutes.

The null and alternative hypothesis are:

$H_0: \mu\geq 300\\\\H_1: \mu$

The level of significance is 0.05.

The test statistic t is

$t=\frac{M-\mu}{s/\sqrt{N}} =\frac{295-300}{20/\sqrt{500}} =\frac{-5}{0.894} =-5.59$

The degrees of freedom are

$df=n-1=500-1=499$

The P-value for this t=-5.59, with 499 degrees of freedom is P=0. The null hypothesis is rejected.

The sample result is enough evidence to reject the inventor's claim, specially because the large sample size.