Question

A light bulb manufacturer guarantees that the mean life of a certain type of light bulb is at least 720 hours. A random sample of 51 light bulbs as a mean of 701.6 hours with a population standard deviation of 62 hours. At an α=0.05, can you support the company’s claim using the test statistic?

Answer 1

`z=(701.6-720)/((62)/(√(51)))=-2.119`

`p_v =P(Z<-2.119)=0.017`

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is lower then 720 hours at 5% of signficance.

Explanation:

Data given and notation

`\bar X=701.6` represent the sample mean

`\sigma=62` represent the population standard deviation

`n=51` sample size

`\mu_o =720` represent the value that we want to test

`\alpha=0.05` 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 at least 720 hours :

Null hypothesis:
`\mu \geq 720`

Alternative hypothesis:
`\mu < 720`

Since we know the population deviation, 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=(701.6-720)/((62)/(√(51)))=-2.119`

P-value

Since is a left tailed test the p value would be:

`p_v =P(Z<-2.119)=0.017`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is lower then 720 hours at 5% of signficance.