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

Part a

The test statistic and P-value of the test is 1.581 and 0.057.

Part b

The type-II error is 0.0088.

Part c

The power of the test is 0.9912.

Explanation:

The Z− test is a statistical test for the mean of a population. It can be used when
`n \geq 30` or when the population is normally distributed and population standard deviation
`\left( \sigma \right)` is known.

The type-II error is defined as the probability of accepting the null hypothesis when it is actually false. It is denoted by the Greek letter beta
`\left( \beta \right)`.

The power of a test against a specific alternative is the probability that the test will reject
`{H_0}` at a chosen significance level
`\alpha` when the specified alternative value of the parameter is true.

The formula for the Z− test is,

Here,
`\bar X` represents the sample mean

`\mu` represents the population mean

`\sigma` represents the population standard deviation

`n` represents the sample size

The formula for the type-II error is,
`\beta = P\left( {{\rm{Accepting}}{H_0}|\mu = {\mu _0}} \right)`

The formula for the power is,
`{\rm{Power}} = 1 - \beta`

Here,
`\beta` represents the type-II error

a)

The sample size is
`n = 10`

The population standard deviation is,
`\sigma = 20`

The sample average is
`\bar X = 185`

Null hypothesis
`{H_0}:\mu = 175[/text </p><p>Alternative hypothesis, [tex]{H_0}:\mu > 175{\rm{ }}\left( {{\rm{Right tailed test}}} \right)`

Calculate the test statistic value.

`\begin{array}{c}\\Z = \frac{{\bar X - \mu }}{{\sigma /\sqrt n }}\\\\ = \frac{{185 - 175}}{{20/\sqrt {10} }}\\\\ = {\rm{1}}.{\rm{581139}}\\\\ = {\rm{1}}.{\rm{581 }}\left( {{\rm{Round to 3 decimal place}}} \right)\\\end{array}`

Calculate the P− value of the hypothesis test.

`\begin{array}{c}\\P - {\rm{value}} = P\left( {Z > {Z_{{\rm{statistic}}}}} \right)\\\\ = 1 - P\left( {Z \le 1.581} \right)\\\\ = 1 - \left( {{\rm{ = NORMSDIST(1}}{\rm{.581)}}} \right){\rm{ }}\left( {{\rm{Use MS Excel}}} \right)\\\\ = 1 - 0.{\rm{943}}0{\rm{61}}\\\\ = 0.0{\rm{56939}}\\\\ = 0.0{\rm{57 }}\left( {{\rm{Round to 3 decimal place}}} \right)\\\end{array}`

b)

Calculate the type-II error.

`\begin{array}{c}\\\beta = P\left( {{\rm{Accepting}}{H_0}|\mu = {\mu _0}} \right)\\\\ = P\left( \mu = 200 \right)\\\\ = P\left( {Z \le \frac{{185 - 200}}{{20/\sqrt {10} }}} \right)\\\\ = P\left( {Z \le - 2.372} \right)\\\\ = \left( {{\rm{ = NORMSDIST( - 2}}{\rm{.372)}}} \right){\rm{ }}\left( {{\rm{Use MS Excel}}} \right)\\\\ = 0.00{\rm{8846}}\\\\ = 0.00{\rm{88 }}\left( {{\rm{Round to 4 decimal place}}} \right)\\\end{array}`

c)

Calculate the power of the test.

`\begin{array}{c}\\{\rm{Power}} = 1 - \beta \\\\ = 1 - 0.0088\\\\ = 0.9912\\\end{array}`