1. You want to determine if classical music and blues music have different effects upon the ability of college students to perform mental tasks requiring concentration.
pt(-2.976,9)+(1-pt(2.976,9))
## [1] 0.01555021

\[ H_0: \mu_{diff}=0\\ H_a: \mu_{diff}\neq0\\ t=\frac{-2.6}{\frac{3.026}{\sqrt{12}}}=-2.976\\ p=P(T\leq-2.976)+P(T\geq2.976)=.0155 \]
We reject \(H_0\) in favor of \(H_a\), students perform differently on multiplication tests depending on whether they are listening to classical music or blues music.

pt(-5.777,39)+(1-pt(5.777,39))
## [1] 1.056836e-06

\[ H_0: \mu_{diff}=0\\ H_a: \mu_{diff}\neq0\\ t=\frac{-3.825}{\frac{4.187}{\sqrt{40}}}=-5.777\\ p=P(T\leq-5.777)+P(T\geq5.777)=0 \]
We reject \(H_0\) in favor of \(H_a\), students perform differently on multiplication tests depending on whether they are listening to classical music or blues music.

pt(-3.403,39)+(1-pt(3.403,39))
## [1] 0.00155375

\[ H_0: \mu_1=\mu_2\\ H_a: \mu_1\neq\mu_2\\\\ SE=\sqrt{\frac{8.37^2}{40}+\frac{10.75^2}{40}}=2.154\\ t=\frac{76.75-84.08}{2.154}=-3.403\\ p=P(T\leq-3.403)+P(T\geq3.403)=0.0015 \]
We reject \(H_0\) in favor of \(H_a\), students perform differently on multiplication tests depending on whether they are listening to classical music or blues music.

  1. Exercise 5.48 on p. 272 of OpenIntro Statistics
  1. Hypothesis test:
    \[ H_0: \mu_L=\mu_H=\mu_J=\mu_B=\mu_G\\ H_a: \text{at least onegroup mean is not equal to the others} \]
  2. Conditions: we will assume independence between the groups, the distributions are normal, groups have similar variability
  3. row 1: 4, 2018.16, 2.188, row 2: 1167, 229.12 row 3: 1171, 269,400.16
  4. We fail to reject \(H_0\), we do not have significant evidence that one of the groups is different.
  1. A local newspaper conducts a phone survey to judge public opinion about a referendum. The referendum passes if a majority of voters vote yes. Of the 980 respondents to the survey, 465 say they plan to vote yes. Is this sufficient information for the newspaper to predict the outcome of the referendum vote, or should they say it is too close to call? Conduct your analysis twice, once using a binomial distribution directly and again using the normal approximation to the binomial distribution. How do the results compare?
1-pbinom(464,980,.5)
## [1] 0.9483821

\[ \pi=\text{probability of voting yes} \\ H_0: \pi=.5\\ H_a: \pi>.5\\ X=\text{votes yes on referendum}=465\\ p=P(X\geq465)=0.948 \]
We fail to reject \(H_0\), not enough evidence that a mojority of voters will vote yes they should say it is too close to call.
With a new hypothesis test:

pnorm(-1.597)
## [1] 0.05513285

\[ H_0: \pi=.5\\ H_a: \pi>.5\\ SE=\sqrt{\frac{.5(1-.5)}{980}}=0.0159\\ z=\frac{\frac{465}{980}-.5}{.0159}=-1.597\\ p=P(z\leq-1.597)=0.0551 \]
We fail to jeject \(H_0\), there is not enough evidence that a mojority of voters will vote yes.

  1. Exercise 6.12 on p. 315 of OpenIntro Statistics
  1. it is a sample statistic because it is a measurement taken from a set of the population
  2. 95% CI:
qnorm(0.975)
## [1] 1.959964

\[ SE=\sqrt{\frac{.48(1-.48)}{1259}}=.01408\\ (.48-1.96*SE,.48+1.96*SE)=(.4524,.5075) \]
We are 95% confident that the proportion of the population who thinks marijuana should be legalized lies in \[(.4524,.5075)\].
c) If the survey was a yes or no response question then 52% responded no and it is a normal distribution.
d) The news piece is not likely but since .5 is in our interval it is possible.

  1. Exercise 6.20 on p. 316 of OpenIntro Statistics
    95% CI with ME=.02: \[ ME=.02=1.96*\sqrt{\frac{.48(1-.48)}{n}}\\ n=\frac{.48(1-.48)}{(\frac{.02}{1.96})^2}=2398 \]
    To limit the margin of error of a 95% CI to 2% we would have to survey 2,398 Americans.

  2. Exercise 6.56
  1. \[ H_0: \pi_{treatment}=\pi_{control}\\ H_a: \pi_{treatment}>\pi_{control}\\ \]
  2. \(\pi_{diff}= \frac{10}{34}-\frac{4}{16}=0.04411\)
  3. p-value is about .5 so we fail to reject \(H_0\). There is not significant evidence that the proportion of people who yawn is differenct for treatment and control groups.