Question

CC Paper Company makes various types of paper products. One of their products is a 30 mils thick paper. In order to ensure that the thickness of the paper meets the 30 mils specification, random cuts of paper are selected and the thickness of each cut is measured. A sample of 256 cuts had a mean thickness of 30.3 mils with a standard deviation of 4 mils. At α = 0.1, test to see if the mean thickness is significantly more than 30 mils.a. Compute the standard error of the mean.b. At 95% confidence using the critical value approach, test to see if the mean thickness is significantly more than 30 mils.c. Show that the p-value approach results in the same conclusion as that of part b.

Answer 1

Answer:

Step-by-step explanation:

Hello!

X: Thickness of a cut of mils paper.

Sample

n= 256 cuts

X[bar]= 30.3 mils

S= 4 mils

a. The hypothesis is that the average thicknes of the paper is greater than 30 mils.

The hypotheses are:

H₀: μ ≤ 30

H₁: μ > 30

α: 0.01

You have no information about the variable distribution so you have to apply the central limit theorem and approximate the distribution of the sample mean to normal. Then you can use the approximation to the standard normal:

Z= (X[bar] - μ)/(S/√n) ≈ N(0;1)

This test is one-tailed to the right and it has only one critical value:

$Z_{1-\alpha }= Z_{1-0.1}= Z_{0.90}= 1.28$

The decision rule is:

If $Z_{H_0}$ ≥ 1.28, you reject the null hypothesis.

If $Z_{H_0}$ < 1.28, you don't reject the null hypothesis.

$Z_{H_0}= \frac{(30.3-30)}{4/\sqrt{256} } = 1.2$

The decision is to nor reject the null hypothesis, then at a significance level of 10% there is not enough evidence to reject the null hypothesis, the average thickness of the paper is less than 30 mils.

*-*

The distribution of the sample mean is:

X[bar]≈ N(μ;σ²/n)

The standard error of the mean is: $\frac{S}{\sqrt{n} }= \frac{4}{\sqrt{256} } = 0.25$

b. Using the same approximation you can calculate the 95% CI:

[X[bar] ± $Z_{1-\alpha /2}$ * $\frac{S}{\sqrt{n} }$]

$Z_{1-\alpha /2}= Z_{1-0.25} = Z_{0.975} = 1.96$

[30.3 ± 1.96 * $\frac{4}{\sqrt{256} }$]

[29.81 ; 30.79]

You can use the confidence interval to decide wheter or nor to reject the null hypothesis. If the interval contains the value of μ stated in the statistical hypothesis, then you don't reject the null hypothesis and if it is not contained, then you reject it. Since the value 30 mils is contained in the 95% confidence interval, you can conclude with a level of significance of 5% that the thickness of the papercuts is on average at most 30 mils.

c. Little reminder: The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).

The region of the p-value is always the same type as the rejection region of the test, in this example both the test and the p-value are one tailed to the right, so you can calculate it as:

P(Z≥1.2) = 1 - P(Z<1.2)= 1 - 0.885= 0.115

The p- value is greater that the alfa level in point b. (5%) so the decision is also to not reject the null hypothesis.

I hope you have a SUPER day!