5.6)
90% confidence interval (65,77) n = 25
sample mean = 71
margin of error = 6
6 = t SE
6 = 1.71 SE
SE = 3.51
5.14)
\(\sigma\) = 250 margin of error \(\leq\) 25
a) 90% confidence interval
SE = \(\sigma\) / \(\sqrt{n}\)
25 = 1.65 x 250/\(\sqrt{n}\)
n = 273
b) For a 99% confidence interval, you would need a larger sample size because you want to be 99% confident that the average SAT score is within the same margin of error as before. The more data you have, the more likely your data will have an average within the desired range.
c) 25 = 2.58 x 250/\(\sqrt{n}\) n = 666
5.20) n = 200
a) No, there is not a clear difference between the average reading and writing scores. The mean difference in the reading and writing scores looks like it is around zero. b) The data comes from a simple random sample and is less than 10% of the overall population. So the observations are independent.
c) Ho: There is not a difference between the average reading and writing scores. \(\mu\)read-\(\mu\)write = 0
HA: There is a difference between the average reading and writing scores. \(\mu\)read-\(\mu\)write \(\neq\) 0
d) Independence: Students were chosen at random and represent less than 10% of the population. The sample size is over 30. The distribution does not look skewed.
e) SE = \(\sigma\) / \(\sqrt{n}\)
SE <- 8.887/sqrt(200)
zscore <- (-.545 - 0)/.628
zscore## [1] -0.8678344p score = 0.38
The p score is much greater than 0.05. We cannot reject the null hypothesis.
f) This could be type 2 error. That is that we did not reject the null hypothesis, but the alternate hypothesis was true. The type 2 error would mean that there is a difference bewteen students’ average reading and writing scores.
g) I would expect a confidence interval for the average distance between reading and writing scores to include zero because the p value was high and we did not reject the null, hypothesis, which states that the difference between reading and writing test scores is zero.
5.32)
Ho: The fuel efficiency for cars with automatic and manual transmissions is the same. \(\mu\)manual-\(\mu\)automatic = 0
HA: The fuel efficiency for cars with automatic transmission is different from the fuel efficiency of cars with manual transmission.
\(\mu\)manual-\(\mu\)automatic \(\neq\) 0
\(\bar{x}\)manual - \(\bar{x}\)automatic = 19.85 - 16.12 = 3.73
SE = 1.13
T = (point estimate - null value) / SE
T = (3.73 - 0)/1.13 = 3.30
df = 25
p > 0.200
Because p is greater than .05, we fail to reject the null hypothesis. The data does not provide strong evidence for a difference between the average fuel efficiency for cars with automatic transmission and the average fuel efficiency for cars with manual transmission.
5.48) Ho: The average number of hours worked is the same for people who have completed different levels of education.
HA: At least one pair of groups who have completed different levels of education work different average number of hours.
b) Observations must be independent within and across groups. We don’t know how the data was collected so we have no way to assess independence. We would need to assume that the subjects in each group are independent.
The data in each group needs to be nearly normal. For each of the samples, the first and third quartiles are equidistant from the median. This isn’t true for the bachelor’s data, but that data has a large sample size so we don’t need to be as concerned with skew. The variability across groups needs to be about equal. The standard deviations for each group are fairly close.
c)
df = c(4,1167, 1171)
SumSq = c(2006.16, 267.382, 269388.16)
MeanSq = c(501.54, 229, NA)
ed <- data.frame(df, SumSq, MeanSq)
rownames(ed) <- c("degree", "residual", "total")
ed## df SumSq MeanSq
## degree 4 2006.160 501.54
## residual 1167 267.382 229.00
## total 1171 269388.160 NAqf(.9318, 4, 1167)## [1] 2.188931