91Ó°ÊÓ

You manufacture and sell a liquid product whose electrical conductivity is supposed to be 5. You plan to make 6 measurements of the conductivity of each lot of product. If the product meets specifications, the mean of many measurements will be 5. You will therefore test

$\begin{array}{r}{H}_0:\mathrm{μ}=5\\ H_a:\mathrm{μ}≠5\end{array}$

If the true conductivity is 5.1, the liquid is not suitable for its intended use. You learn that the power of your test at the 5% significance level against the alternative $\mathrm{μ}$=5.1 is 0.23.

(a) Explain in simple language what "power =0.23 " means in this setting.

(b) You could get higher power against the same alternative with the same $\mathrm{Î\pm }$by changing the number of measurements you make. Should you make more measurements or fewer to increase power?

(c) If you decide to use $\mathrm{Î\pm }$=0.10 in place of $\mathrm{Î\pm }$=0.05, with no other changes in the test, will the power increase or decrease? Justify your answer.

(d) If you shift your interest to the alternative $\mathrm{μ}$=5.2, with no other changes, will the power increase or decrease? Justify your answer.

Step by step solution

01

Given Information

The electrical conductivity of the liquid product = $5$

$\begin{array}{r}{H}_0:\mathrm{μ}=5\\ H_a:\mathrm{μ}≠5\end{array}$

Significance level =$5\%$

$\mathrm{μ}=5.1$

0.23p =$0.23$

02

Explanation Part (a)

Given,

Significance level =$5\%$

$\begin{array}{c}\mathrm{μ}=5.1\\ p=0.23\end{array}$

"power =$0.23$ " means the probability of rejecting the null hypothesis if it's false

03

Explanation Part (b)

Given,

Significance level =$5\%$

$\begin{array}{r}\mathrm{μ}=5.1\\ p=0.23\end{array}$

We could get higher power against the same alternative with the same$\mathrm{Î\pm }$by changing the number of measurements you make hence we should make more measurements as more information will be needed.

04

Explanation Part (c)

As the probability of making type I error increases the probability of making a type II error decreases.

We know,

$\text{Power}=1−\mathrm{β}$(the probability of making a type II error)

Hence as the probability of making a type II error decreases the power increases.

05

Explanation Part (d)

Given,

Shifting the alternative to $\mathrm{μ}=5.2$

As per the null hypothesis,

$H_0:\mathrm{μ}=5$

The alternative is greater than the null hypothesis, Hence it means the power has increased.

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