Q 91. A company that manufactures clas... [FREE SOLUTION]

91影视

The Practice of Statistics for AP Examination

Daren Starnes, Josh Tabor

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 837 pages

ISBN: 9781319113339

Chapter 9: Q 91. (page 609)

A company that manufactures classroom chairs for high school students
claims that the mean breaking strength of the chairs is 300 pounds. One of the chairs collapsed beneath a 220-pound student last week. You suspect that the manufacturer is exaggerating the breaking strength of the chairs, so you would like to perform a test of $H_{0} : 渭 = 300 H_{a} : 渭 < 300$ where 渭 is the true mean breaking strength of this company鈥檚 classroom chairs.
a. The power of the test to detect that 渭=294 based on a random sample of 30
chairs and a significance level of 伪=0.05 is 0.71. Interpret this value.
b. Find the probability of a Type I error and the probability of a Type II error for the test in part (a).
c. Describe two ways to increase the power of the test in part (a).

Short Answer

Expert verified

Part (a) If the true mean breaking strength of this company鈥檚 classroom chairs is 2 $94$ pounds, then have a $71 %$ probability that find the convincing proof to help the alternative hypothesis $H_{1} : 渭 < 300$

Part (b) $P (T y p e I e r r o r) = 伪 = 0.05 = 5 % P (T y p e I I e r r o r) = 0.29 = 29 %$

Part (c) Increase the significance level.

Increase sample size.

Making the alternative mean more extreme.

Step by step solution

Part (a) Step 1: Given information

$H_{0} : 渭 = 300 H_{a} : 渭 < 300$

$渭_{A} = 294 n = 30 伪 = 0.05 P o w e r = 0.71 = 71 %$

Part (a) Step 2: Explanation

When the alternative hypothesis is true, the power is the probability of rejecting the null hypothesis. If the genuine mean breaking strength of this company's classroom chairs is $294$ pounds, then the alternative hypothesis has a $71$ percent chance of being supported $H_{1} : 渭 < 300$

Part (b) Step 1: Explanation

The significance level determines the likelihood of a type I error:

$P (T y p e I e r r o r) = 伪 = 0.05 = 5 %$

The power reduces the probability of a type II error by one.

$P (T y p e I I e r r o r) = 1 鈭� P o w e r = 1 - 0.71 = 0.29 = 29 %$

Part (c) Step 1: Explanation

When the alternative hypothesis is true, the power is the probability of rejecting the null hypothesis.

Increase the power by:

Boost the level of significance (because this increases the probability of making a Type I error and decreases the probability of making a Type II error; since the power is $1$ decreased by the probability of making a Type II error, the power increases).

Making the alternative more harsh entails reducing the size of the alternative (since more extreme alternatives are easier to prove)