Question

A consumer products company is formulating a new shampoo and is interested in foam height (in mm). Foam height is approximately normally distributed and has a standard deviation of 20 mm. The company wishes to test H0: μ = 175 mm versus H1: μ > 175 mm, using a random sample of n = 10 samples.(a) Find P-value if the sample average is = 185? Round your final answer to 3 decimal places.(b) What is the probability of type II error if the true mean foam height is 200 mm and we assume that α = 0.05? Round your intermediate answer to 1 decimal place. Round the final answer to 4 decimal places.(c) What is the power of the test from part (b)? Round your final answer to 4 decimal places.

Answer 1

Answer:

a) 0.057

b) 0.5234

c) 0.4766

Step-by-step explanation:

a)

To find the p-value if the sample average is 185, we first compute the z-score associated to this value, we use the formula

$z=\frac{\bar x-\mu}{\sigma/\sqrt N}$

where

$\bar x=mean\; of\;the \;sample$

$\mu=mean\; established\; in\; H_0$

$\sigma=standard \; deviation$

N = size of the sample.

So,

$z=\frac{185-175}{20/\sqrt {10}}=1.5811$

$\boxed {z=1.5811}$

As the sample suggests that the real mean could be greater than the established in the null hypothesis, then we are interested in the area under the normal curve to the right of  1.5811 and this would be your p-value.

We compute the area of the normal curve for values to the right of  1.5811 either with a table or with a computer and find that this area is equal to 0.0569 = 0.057 rounded to 3 decimals.

So the p-value is  

$\boxed {p=0.057}$

b)

Since the z-score associated to an α value of 0.05 is 1.64 and the z-score of the alternative hypothesis is 1.5811 which is less than 1.64 (z critical), we cannot reject the null, so we are making a Type II error since 175 is not the true mean.

We can compute the probability of such an error following the next steps:

Step 1

Compute $\bar x_{critical}$

$1.64=z_{critical}=\frac{\bar x_{critical}-\mu_0}{\sigma/\sqrt{n}}$

$\frac{\bar x_{critical}-\mu_0}{\sigma/\sqrt{n}}=\frac{\bar x_{critical}-175}{6.3245}=1.64\Rightarrow \bar x_{critical}=185.3721$

So we would make a Type II error if our sample mean is less than 185.3721.  

Step 2

Compute the probability that your sample mean is less than 185.3711  

$P(\bar x < 185.3711)=P(z< \frac{185.3711-185}{6.3245})=P(z$

So, the probability of making a Type II error is 0.5234 = 52.34%

c)

The power of a hypothesis test is 1 minus the probability of a Type II error. So, the power of the test is

1 - 0.5234 = 0.4766