1. In the test statistic, we use the population value, p₀, that is specified in H₀, but in the confidence interval, we do not have a pop. value, so we use a sample value, p hat, instead.

2. The power of the test is the probability of rejection H₀ when it is false. 
 
 

3. T

4. T

5. F You must have equal sample sizes since the test is based on the differences between the pairs. 
 
 

6. T

7. F If the Ha is µ< µ_o then you would expect t to be negative and if it is negative enough, Ho may be rejected.

8. F No, for example, if the sample mean is LESS than the pop. mean and the two sided test is significant, but the one-sided test is µ>µ_o, then the null hypothesis would not be rejected.
 

II. Problems.
 

1. a) n=10 is small, so must assume that the satisfaction scores are normally distributed.

b) xbar = 1258/10 = 125.8 s =√94.62 = 9.7274 t(19,99%) = 3.25 
 

125.8 +/- 3.25 (9.7274/√10) or 125.8 +/- 3.25(3.0761) or 125.8 +/-  9.997    (115.80, 135.80)
 

c)We are 99% confident that the interval from 115.80 to 135.80 contains the population mean satisfaction score for all passenger cars in 2003.
 

d) No, it is not likely the mean satisfaction has increased from 2002 since the 2002 value, 119, is in the confidence interval for 2003. (We would not have rejected a 2-sided H₀ with α=.01).
 

2. a) Ho: The population proportion of women who think shopping is pleasant is equal to the population proportion of men who do. Ho: p(w) = p(m)
 

Ha: The population proportion of women who think shopping is pleasant differs from the population proportion of men who do. Ha: p(w) ≠ p(m)

b)55, 41, 74, 27 are all > 5 so CLT holds for phat _w and phat_m.

  z = √ 5.557 = 2.357 subtracting women - men.
 

c) p-value = .018 from the SPSS printout or 2P(z>2.36) = 2(.0091) = .0182. 
 

d) .018> .01 so do not reject Ho. There is not sufficient sample evidence to conclude that the population proportion of women who think shopping is pleasant differs from the population proportion of men who do.
 

3. a) Ho: The population mean semester grade average for students with less than 2 years of algebra is equal to that for students with at least 2 years of algebra.
 

Ha: The population mean semester grade average for students with less than 2 years of algebra is less than that for students with at least 2 years of algebra..

Ho: µ(<2yrs) = µ(≥2yrs) Ha: µ(<2yrs)< µ(≥2yrs)

b) n₁=6 and n₂=9 are small, so must assume that the semester grade averages in both pop. are normally distributed. The boxplots have no outliers, the medians are not terribly off center and the whiskers are about equal so the assumption of normality is ok. Must also assume that σ(<2yrs) = σ(≥2yrs). This holds since Levene's test of equal σ's is not rejected, p=.929 > .05. 
 

c) From SPSS, equal t =-1.709 
 

c) Since t is negative and the sample mean for <2yrs is 63.5 which is less than the mean for 2+ years, 74.8, we find the p-value from SPSS-- p =.111 /2 =.0555.
 

d) There is a 5.55% chance of getting a sample mean difference in semester grade averages of -11.28 points or less between the <2 yrs of algebra group and the 2+ yrs of algebra group if the true mean difference is zero.
 

e) .0555 > .05 so do not reject Ho. There is not sufficient evidence to conclude that the population mean semester grade average for students with less than 2 years of algebra is less than that for students with at least 2 years of algebra.
 

4.a) n = (1.645/.06)²(.5)(.5)= 187.92 or n=188

b) X = 119 and n-X = 188 - 119 = 59 are both > 15 so the CLT for holds.

c) z = 1.645 = 119/188 = .6330 so .6330 +/- 1.645 √(.6330)(.3670)/188 

or .633  +/- 1.645 (.0352) or .633  +/- .0578 or (.5752, .6908)
 

d) We are 90% confident that the interval from .575 to .691 contains the population proportion of viewers who tuned their TVs to the premiere of a new hospital drama show.
 

e) We wanted a margin of error of .06 and we got .0578 which is less than .06 so yes, the margin of error was achieved with a sample size of 188.