Question

A new process for producing a type of resin is supposed to have a mean cycle time of 3.5 hours

per batch. Six batches are produced, and their cycle times, in hours, were:

3.45 3.47 3.57 3.52 3.40 3.63

Can you conclude that the mean cycle time is greater than 3.5 hours?
a. Use the critical value approach.
b. Use the p-value approach

Answer 1

a) Assuming 95% of confidence and
`\alpha=0.05` we can use the t distribution with 5 degrees of freedom in order to calculate a critical value that accumulates 0.05 of the area on the right tail of the distribution. We can use excel and the code to do this is given by: "=T.INV(1-0.05,5)". And we got the critical value
`t_(\alpha/2)=2.015`.

Since our calculates value < critical value. We fail to reject the null hypothesis, and we can say that at 5% of significance we don't have enough evidence to conclude that the true mean is highr than 3.5 hours.

b)
`p_v =P(t_(5)>0.204)=0.423`

If we compare the p value and a significance level assumed
`\alpha=0.05` we see that
`p_v>\alpha` so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean for the time is not significantly higher than 3.5 hours at 5% of significance.

Explanation:

Data given and notation

We can calculate the sample mean and standard deviation with these formulas:

`\bar X =(\sum_(i=1)^n X_i)/(n)`

`s =\sqrt{(\sum_(i=1)^n (X_i -\bar X)^2)/(n-1)}`

The results obtained are given below:

`\bar X=3.507` represent the sample mean

`s=0.084` represent the standard deviation for the sample

`n=6` sample size

`\mu_o =3.5` 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)

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the mean producing a type of resin is higher than 3.5 hours, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 3.5`

Alternative hypothesis:
`\mu > 3.5`

Compute the test statistic

We don't know the population deviation, so for this case 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".

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

`t=(3.507-3.5)/((0.084)/(√(6)))=0.204`

Now we need to find the degrees of freedom for the t distirbution given by:

`df=n-1=6-1=5`

What do you conclude?

a. Use the critical value approach.

Assuming 95% of confidence and
`\alpha=0.05` we can use the t distribution with 5 degrees of freedom in order to calculate a critical value that accumulates 0.05 of the area on the right tail of the distribution. We can use excel and the code to do this is given by: "=T.INV(1-0.05,5)". And we got the critical value
`t_(\alpha/2)=2.015`.

Since our calculates value < critical value. We fail to reject the null hypothesis, and we can say that at 5% of significance we don't have enough evidence to conclude that the true mean is highr than 3.5 hours.

b. Use the p-value approach

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

`p_v =P(t_(5)>0.204)=0.423`

If we compare the p value and a significance level assumed
`\alpha=0.05` we see that
`p_v>\alpha` so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean for the time is not significantly higher than 3.5 hours at 5% of significance.