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The Practice of Statistics for AP Examination

Daren Starnes, Josh Tabor

$Math Studyset Vaia Explanations$ Math

6th Edition · 837 pages

ISBN: 9781319113339

Chapter 10: Q. AP3.25 (page 704)

Suppose the null and alternative hypothesis for a significance test are defined as
H0: μ= $40$ $\frac{305}{1526}$ $= 0.200 = 20.0 %$ H0 : μ= $40$
Ha: μ< $40$ $\frac{305}{1526} = 0.200 = 20.0 %$ Ha : μ< $40$
Which of the following specific values for Ha will give the highest power? a. μ= $38$ $\frac{305}{1526} = 0.200 = 20.0 %$ $μ = 38$
b. μ= $39$ $\frac{305}{1526} = 0.200 = 20.0 % μ = 39$
c. μ= $41$ $\frac{305}{1526} = 0.200 = 20.0 % μ = 41$
d. μ= $42$ $\frac{305}{1526} = 0.200 = 20.0 % μ = 42$
e. μ= $43$ $\frac{305}{1526} = 0.200 = 20.0 % μ = 43$

Short Answer

Expert verified

The correct option is - (a) $μ = 38$ will give the highest power .

Step by step solution

01

Given Information

We are given a null hypothesis and alternate hypothesis for a given significance test . We need to find which will give the highest power .

02

Explanation

The null and alternate hypothesis give the highest power when it is less than and furthest from $40$ . So, the given furthest value is $38$ . Hence it will give the highest power .

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Most popular questions from this chapter

Which of the following will increase the power of a significance test?

a. Increase the Type II error probability.

b. Decrease the sample size.

c. Reject the null hypothesis only if the P-value is less than the significance level.

d. Increase the significance level $α$ .

e. Select a value for the alternative hypothesis closer to the value of the null hypothesis.

A random sample of $200$ New York State voters included $88$ Republicans, while a random sample of $300$ California voters produced $141$ Republicans. Which of the following represents the $95 %$ confidence interval for the true difference in the proportion of Republicans in New York State and California?

a. $(0.44 - 0.47) \pm 1.96 ((0.44) (0.56) + (0.47) (0.53) 200 + 300)$

b. $(0.44 - 0.47) \pm 1.96 ((0.44) (0.56) 200 + (0.47) (0.53) 300)$

c. ( $0.44 - 0.47) \pm 1.96 (0.44) (0.56) 200 + (0.47) (0.53) 300$

d. $(0.44 - 0.47) \pm 1.96 (0.44) (0.56) + (0.47) (0.53) 200 + 300$

e. $(0.44 - 0.47) \pm 1.96 (0.45) (0.55) 200 + (0.45) (0.55) 300$

Facebook As part of the Pew Internet and American Life Project, researchers conducted two surveys. The first survey asked a random sample of $1060$ U.S. teens about their use of social media. A second survey posed similar questions to a random sample of $2003$ U.S. adults. In these two studies, $71.0 %$ of teens and $58.0 %$ of adults used Facebook. Let plocalid="1654194806576" $T^{\frac{305}{1526} = 0.200 = 20.0 %}$ pT= the true proportion of all U.S. teens who use Facebook and p $A^{\frac{305}{1526} = 0.200 = 20 %}$ pA = the true proportion of all U.S. adults who use Facebook. Calculate and interpret a $99 %$ confidence interval for the difference in the true proportions of U.S. teens and adults who use Facebook.

Coaching and SAT scores Let’s first ask if students who are coached increased their scores significantly, on average.

a. You could use the information on the Coached line to carry out either a two-sample t test comparing Try 1 with Try 2 or a paired t test using Gain. Which is the correct test? Why?

b. Carry out the proper test. What do you conclude?

Researchers suspect that Variety A tomato plants have a different average yield than Variety B tomato plants. To find out, researchers randomly select $10$ Variety A and $10$ Variety B tomato plants. Then the researchers divide in half each of $10$ small plots of land in different locations. For each plot, a coin toss determines which half of the plot gets a Variety A plant; a Variety B plant goes in the other half. After harvest, they compare the yield in pounds for the plants at each location. The $10$ differences (Variety A − Variety B) in yield are recorded. A graph of the differences looks roughly symmetric and single-peaked with no outliers. The mean difference is ${\bar{x}}_{A - B}$ $= 0.34$ $\frac{305}{1526} = 0.200 = 20 % {\bar{x}}_{A - B} = 0.34$ and the standard deviation of the differences is s A-B $= 0.83$ $\frac{305}{1526} = 0.200 = 20 % = s_{A - B} = 0.83$ .Let $μ_{A - B} = \frac{305}{1526} = 0.200 = 20 %$ μA−B = the true mean difference (Variety A − Variety B) in yield for tomato plants of these two varieties.

A 95% confidence interval for $μ_{A - B} \frac{305}{1526} = 0.200 = 20 % μ_{A - B}$ is given by

a. $0.34 \pm 1.96 (0.83) \frac{305}{1526} = 0.200 = 20 % 0.34 \pm 1.96 (0.83)$

b. $0.34 \pm 1.96 (0.8310) \frac{305}{1526} = 0.200 = 20 % 0.34 \pm 1.96 (0.8310)$

c. $0.34 \pm 1.812 (0.8310) \frac{305}{1526} = 0.200 = 20 % 0.34 \pm 1.812 (0.8310)$

d. $0.34 \pm 2.262 (0.83) \frac{305}{1526} = 0.200 = 20 % 0.34 \pm 2.262 (0.83)$

e. $0.34 \pm 2.262 (0.8310) \frac{305}{1526} = 0.200 = 20 % 0.34 \pm 2.262 (0.8310)$

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