5.6

i) mean = 71

ii) margin of error = 6

iii) s = 17.5644028

5.14

Deriving formula:
\[t\times SE=25\]
\[SE = 25/t\]

\[s/\sqrt { n } \quad =\quad (25/t)\]
\[n=(s^{ 2 }\times t^{ 2 })/25^2\]

To not have two unknowns with one equation, assume sample is large enough to substitute Z for t

a) With 90% confidence interval \[((250^{ 2 })\times 1.645^{ 2 })/25^2\] = 270.6025

b) Luke’s sample will need to be larger because from the formula we can see that sample size needed is proportional to the square of the test statistic.

c) With 99% confidence interval \[(250^{ 2 }\times 2.575^{ 2 })/25^2\] = 663.0625

5.20

a) There is a difference in medians, but I’d say there isn’t a clear difference as the IQR has about the same bounds and the histogram is centered around 0.

b) The scores are not likely independant. Students who do well in one area are likely to do well in the other area. Scores between students are likely independant, so we can proceed with the test.

C) Null Hypothesis: The average differenc between an individual’s reading and writing score is 0 Alternative Hypothesis: The average difference between an individuals reading and writing score is not 0

d) I would say the Normality of the distribution condition is satisfyed based on the historgram, along with the independance of observations based on the fact that the sample is random and likely not more than 10% of the population. We can still proceed even though each observation isn’t independant of itself.

e) SE = S/sqrt(200) = 0.6284058 t = 0.867274 There is not convincing evidence of a difference.

f) We could have possibly made a type II error, since we did not reject the null hypothesis.

g) 0 falls within less than 1 SE of the observed mean, so it would fall within any reasonable confidence interval, 90% for example

5.32

I would choose not to pool standard deviation. Because standards these days are likely to be muscle cars, while automatics could come from a wide range.

SE <- sqrt(3.58^2/26 + 4.51^2/26)
t <- (19.85 - 16.12)/SE
t
## [1] 3.30302

We use a two tailed test since the experiment is just for a difference and the critical value is 2.0555 at 5% significance. The t score is higher than this so we reject the null hypothesis that the means are equal.

5.48

a) Null hypothesis: there is no difference in mean hours worked among the five groups Alternative hypothesis: There is a difference in mean hours worked among the five groups

b) There is independance in observations within and between the groups. The groups have approximately the same spread and are normally distributed within the groups.

c)

anova output

anova output

d) We did not choose a level of significance, but .05 seems typical for F tests, so we can not reject the null hypothesis at that level.