Question

A new centrifugal pump is being considered for an application involving the pumping of ammonia. The specification is that the flow rate be more than 5 gallons per minute (gpm). In an initial study, eight runs were made. The average flow rate was 6.5gpm and the standard deviation was 1.9 gpm. If the mean flow rate is found to meet the specification, the pump will be put into service.

1. State the appropriate null and alternate hypotheses

2. Find the P-value

3. Should the pump be put into service? Explain.

Answer 1

1) We need to conduct a hypothesis in order to check if the true mean is higher than 5 gpm, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 5`

Alternative hypothesis:
`\mu > 5`

2)
`df=n-1=8-1=7`

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

`p_v =P(t_((7))>2.233)=0.0304`

3) For this case is we use a significance level of 1% or 99% of confidencewe see that
`p_v >\alpha` and we don't have enough evidence to conclude that the specification is satified. But if we use a value of significance
`\alpha=0.05` or 95% of confidence we see that
`p_v <\alpha` and we have enough evidence to conclude that the specification is satisfied.

Explanation:

Data given and notation

`\bar X=6.5` represent the sample mean

`s=1.9` represent the sample standard deviation

`n=8` sample size

`\mu_o =5` represent the value that we want to test

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

t would represent the statistic (variable of interest)

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

Part 1: State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is higher than 5 gpm, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 5`

Alternative hypothesis:
`\mu > 5`

If we analyze the size for the sample is <30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

t-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:

`t=(6.5-5)/((1.9)/(√(8)))=2.233`

Part 2: P-value

The first step is calculate the degrees of freedom, on this case:

`df=n-1=8-1=7`

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

`p_v =P(t_((7))>2.233)=0.0304`

Part d: 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 height of men actually its significant higher compared to the height of men in 1960 at 1% of signficance.

Part 3

For this case is we use a significance level of 1% or 99% of confidencewe see that
`p_v >\alpha` and we don't have enough evidence to conclude that the specification is satified. But if we use a value of significance
`\alpha=0.05` or 95% of confidence we see that
`p_v <\alpha` and we have enough evidence to conclude that the specification is satisfied.