AP Stats - Chapter 9 (9.1, 9.2 & 9.3 Combined)

Experiments on learning in animals sometimes

measure how long it takes mice to find their way

through a maze. The mean time is 18 seconds for

one particular maze. A researcher thinks that a loud

noise will cause the mice to complete the maze

faster. She measures how long each of 10 mice takes

with a noise as stimulus. The appropriate hypotheses

for the significance test are

• $H_o:\ \mu\ =\ 18;\ H_a:\ \mu\ne18$

• $H_o:\ \mu\ =\ 18;\ H_a:\ \mu>18$

• $H_o:\ \mu\ <\ 18;\ H_a:\ \mu=18$

• $H_o:\ \mu\ =\ 18;\ H_a:\ \mu<18$

• $H_o:\ \overline{x}=\ 18;\ H_a:\ \overline{x}\ <18$

Members of the city council want to know if a majority of city residents supports a 1% increase in the sales tax to fund road repairs. To investigate, they survey a random sample of 300 city residents and use the results to test the following hypotheses:

 $H_o:p\ =\ .50$  
 $H_a:p>.50$  
where p is the proportion of all city residents who support a 1% increase in the sales tax to fund road repairs. 

A Type I error in the context of this study occurs if the city council

• finds convincing evidence that a majority of residents supports the tax increase, when in reality there isn’t convincing evidence that a majority supports the increase.

• finds convincing evidence that a majority of residents supports the tax increase, when in reality at most 50% of city residents support the increase.

• finds convincing evidence that a majority of residents supports the tax increase, when in reality more than 50% of city residents do support the increase.

• does not find convincing evidence that a majority of residents supports the tax increase, when in reality more than 50% of city residents do support the increase.

• does not find convincing evidence that a majority of residents supports the tax increase, when in reality at most 50% of city residents do support the increase.

Members of the city council want to know if a majority of city residents supports a 1% increase in the sales tax to fund road repairs. To investigate, they survey a random sample of 300 city residents and use the results to test the following hypotheses:

 $H_o:\ p\ =\ 0.50$  
 $H_a:\ p\ >0.50$  
  
where p is the proportion of all city residents who support a 1% increase in the sales tax to fund road repairs. 

In the sample, p-hat = 158/300 = 0.527. The resulting
P-value is 0.18. What is the correct interpretation of
this P-value?

• Only 18% of the city residents support the tax increase.

• There is an 18% chance that the majority of residents

  supports the tax increase.

• Assuming that 50% of residents support the tax

  increase, there is an 18% probability that the sample

  proportion would be 0.527 or higher by chance alone.

• Assuming that more than 50% of residents support the

  tax increase, there is an 18% probability that the sample

  proportion would be 0.527 or higher by chance alone.

• Assuming that 50% of residents support the tax

  increase, there is an 18% chance that the null

  hypothesis is true by chance alone.

Members of the city council want to know if a majority of city residents supports a 1% increase in the sales tax to fund road repairs.To investigate, they survey a random sample of 300 city residents and use the results to test the following hypotheses:
 $H_o:\ p\ =\ 0.50$  
 $H_a:\ p\ >0.50$  

where p is the proportion of all city residents who support a 1% increase in the sales tax to fund road repairs.

Based on the P-value in Exercise 27, which of the following would be the most appropriate conclusion?

• Because the P-value is large, we reject  $H_o$  . We have convincing evidence that more than 50% of city residents support the tax increase.

• Because the P-value is large, we fail to reject  $H_o$  . We have convincing evidence that more than 50% of city residents support the tax increase.

• Because the P-value is large, we reject  $H_o$  . We have convincing evidence that at most 50% of city residents support the tax increase.

• Because the P-value is large, we fail to reject  $H_o$  . We have convincing evidence that at most 50% of city residents support the tax increase.

• Because the P-value is large, we fail to reject  $H_o$  . We do not have convincing evidence that more than 50% of city residents support the tax increase.

After once again losing a football game to the

archrival, a college’s alumni association conducted

a survey to see if alumni were in favor of firing the

coach. An SRS of 100 alumni from the population

of all living alumni was taken, and 64 of the alumni

in the sample were in favor of firing the coach. Sup-

pose you wish to see if a majority of living alumni

are in favor of firing the coach. The appropriate test

statistic is

• $z=\frac{0.64-0.50}{\sqrt{\frac{0.64\left(0.36\right)}{100}}}$

• $t=\frac{0.64-0.50}{\sqrt{\frac{0.64\left(0.36\right)}{100}}}$

• $z=\frac{0.64-0.50}{\sqrt{\frac{0.5\left(0.5\right)}{100}}}$

• $z=\frac{0.64-0.50}{\sqrt{\frac{0.64\left(0.36\right)}{64}}}$

• $z=\frac{0.50-0.64}{\sqrt{\frac{0.64\left(0.36\right)}{100}}}$

Which of the following is not a condition for

performing a significance test about a population

proportion p?

• The data should come from a random sample or

  randomized experiment.

• Both $np_o\ and\ n\left(1-p_o\right)$ should be at least 10.

• If you are sampling without replacement from a finite population, then you should sample no more than 10% of the population.

• The population distribution should be approximately Normal, unless the sample size is large.

• All of the above are conditions for performing a significance test about a population proportion.

Given a test with hypotheses
 $H_o:\ p=0.4\ \ and\ \ H_a:\ p\ne0.4$  ,

the test statistic is z = 2.43.  This test is

•  $Not\ significant\ at\ \alpha=0.05\ or\ at\ \ \alpha=0.01$ 

•  $Significant\ at\ \alpha=0.05\ but\ not\ at\ \ \alpha=0.01$  

•  $Significant\ at\ \alpha=0.01\ but\ not\ at\ \alpha=0.05$  

•  $Significant\ at\ both\ \alpha=0.05\ and\ \ \alpha=0.01$  

• inconclusive because we don't know the value of p-hat

Which of the following 95% confidence intervals
would lead us to reject 
 $H_o:\ p=0.3$ in favor of
  $H_a:\ p\ne0.3$  at the 5% significance level

• (0.19, 0.27)

• (0.24, 0.30)

• (0.27, 0.31)

• (0.29, 0.38)

• None of these

The reason we use t procedures instead of z procedures when carrying out a test about a population mean is that

• z requires that the sample size be large.

• z requires that you know the population standard deviation $\sigma$

• z requires that the data come from a random sample or randomized experiment.

• z requires that the population distribution be perfectly Normal.

• z can only be used for proportions.

You are testing $H_o:\ \mu=10$  against $H_a:\ \mu<10$  based on an SRS of 20 observations from a Normal population. The t statistic is t = −2.25. The P-value

• falls between 0.01 and 0.02.

• falls between 0.02 and 0.04.

• falls between 0.04 and 0.05.

• falls between 0.05 and 0.25.

• is greater than 0.25.