Question

The filling machine for a production operation must be adjusted if more than 8% of the items being produced are underfilled. A random sample of 80 items from the day’s production contained nine underfilled items. Does the sample evidence indicate that the filling machine should be adjusted? Use a level of significance of 0.01.

Answer 1

The sample evidence doesn't indicate that the filling machine should be adjusted.

Explanation:

First, we need to define the null and alternative hypothesis as:

`H0: p=0.08\\H1: p>0.08`

Where p is the proportion of the items produced.

Then, we can find the statistic using the following equation:

`z=\frac{p'-p}{\sqrt{(p(1-p))/(n) } }`

Where p' is the proportion of the sample and n is the size of the sample. Replacing p' by 0.1125 (9/80=0.1125) and n by 80, we get:

`z=\frac{0.1125-0.08}{\sqrt{(0.08(1-0.08))/(80) } }=1.07`

Finally, we can calculated the p-value as:

`p-value=P(z>1.07)=0.1423`

So, p-value is greater than the level of significance. It means that we can the null hypothesis and the sample evidence doesn't indicate that the filling machine should be adjusted.

Answer 2

`z=\frac{0.1125 -0.08}{\sqrt{(0.08(1-0.08))/(80)}}=1.071`

The p value is given by:

`p_v =P(z>1.071)=0.142`

For this case the p value is higher than the significance level of 0.01 so then we have enough evidence to FAIl to reject the null hypothesis and we can't conclude that the true proportion is significanctly higher than 8%

Explanation:

Information given

n=80 represent the random sample taken

X=9 represent the number fo items undefilled

`\hat p=(9)/(80)=0.1125` estimated proportion of underfilled items

`p_o=0.08` is the value that we want to test

`\alpha=0.01` represent the significance level

z would represent the statistic

`p_v` represent the p value

Hypothesis to test

We want to test if the true proportion of items underfilled is higher than 0.08 and we got.:

Null hypothesis:
`p\leq 0.08`

Alternative hypothesis:
`p > 0.08`

The statistic is given by:

`z=\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}}` (1)

Replacing the info given by:

`z=\frac{0.1125 -0.08}{\sqrt{(0.08(1-0.08))/(80)}}=1.071`

The p value is given by:

`p_v =P(z>1.071)=0.142`

For this case the p value is higher than the significance level of 0.01 so then we have enough evidence to FAIl to reject the null hypothesis and we can't conclude that the true proportion is significanctly higher than 8%