Question

: An office building in New York uses thousands of fluorescent light bulb each year. The brand of bulb it currently uses has a mean life of 900 hours. A manufacturer claims that its new brand of bulbs, which cost the same as the brand the office building currently uses, has a mean life of more than 900 hours. The university has decided to purchase the new brand if, when tested, the test evidence supports the manufacturer’s claim at the .05 significance level. Suppose 64 bulbs and the sample has a mean of 920 hours and a standard deviation of 80 hours.

What are the null hypothesis and the alternative hypothesis?
Should the office building purchase the new brand of florescent bulbs? (i.e. test your hypothesis using a z-test)

Answer 1

Null hypothesis:
`\mu \leq 900`

Alternative hypothesis:
`\mu > 900`

`z=(920-900)/((80)/(√(64)))=2`

If we use the z distribution the p value would be:

`p_v =P(Z>2)=0.02275`

And if we use the t distribution the first step is calculate the degrees of freedom, on this case:

`df=n-1=64-1=63`

Since is a one side right tailed test the p value would be:

`p_v =P(t_((63))>2)=0.0249`

If we compare the p values obtained and the significance level given
`\alpha=0.05` for both we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can conculde that the mean for the new brand it's significantly higher than 900 hours.

Explanation:

Data given and notation

`\bar X=920` represent the mean height for the sample

`s=80` represent the sample standard deviation for the sample

`n=64` sample size

`\mu_o =900` 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 higher than 900 hours, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 900`

Alternative hypothesis:
`\mu > 900`

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, but the problem says that we need to use the z test on this case 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=(920-900)/((80)/(√(64)))=2`

P-value

If we use the z distribution the p value would be:

`p_v =P(Z>2)=0.02275`

And if we use the t distribution the first step is calculate the degrees of freedom, on this case:

`df=n-1=64-1=63`

Since is a one side right tailed test the p value would be:

`p_v =P(t_((63))>2)=0.0249`

Conclusion

If we compare the p values obtained and the significance level given
`\alpha=0.05` for both we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can conculde that the mean for the new brand it's significantly higher than 900 hours.