Problem 4.4
a) The point est. for the average height is 171.1 cm. 170.3 cm is the point estimate of the median.
b) The point estimate for the standard deviation of the heights is 9.4 cm. The IQR is 177.8-163.8 = 14. c) A person who is 180 cm tall is just under 1 sd from the mean. That’s not very unusual. A person who is 155cm is ~ 1.7 sd from the mean. That’s my wife’s height. I’m not going to call it unusual. (This data includes males.)
d) It would be quite surprising if the mean and sd of the disribution from another sample is the same.
e) The standard error is 9.4/(sqrt(507))=.417
Problem 4.14
a) True. Enough people were surveyed to believe our conf. interval.
b) False. With a large enough sample, we can accept the CLT and the CI.
c) False. We expect this to prove true, but don’t know.
d) True.
e) True. The lower the percent sure we need to be, the narrower the range will be.
f) False. SE is proportional to 1/sqrt(N) g) True. It’s half the CI.
Problem 4.24
a) The sampling appears to e random, with iid. It doesn’t appear to be incredibly skewed, but a little bit.
b) Pr (30.69-32)/(4.39/6)) = pr (Z < -1.79) = .0367 Given our null hyposthesis, there is a 3.67% of getting the distribution we observed. The H0 is rejected at a conf. level of .1
c) The p-value, .0367 is less than .1, our test threshold. d) The ci is 30.69-(1.645)(4.31/6) = 29.51 to 30.69-(1.645)(4.31/6) = 31.87 e) The hypothesis test and the con???dence interval agree; both would reject it. Both are different interpretations of similar information.
Problem 4.26
a) (118.2-100)/(6.5/6) = 16.8 That’s a very large Z-score. The H0 is rejected at a significance level of .1
b) The confidence interval is 116.42-119.98. (118.2 +/- 1.645 (6.5/6) ) c) Both tests agree. The 90% confidence interval is very far from the larger population mean.
Problem 4.34
A sampling distribution is an empirical distribution drawn from a random selection from a larger sample. The selection is run many times. We can use the CLT to describe it because it involves many samples drawn randomly from a sampling. It is an asymptotically unbiased estimator for the population mean and sd.
Problem 4.40
a) 1-Phi((10500-9000)/1000) = 1-Phi(1.5)=1-.9332= 6.68%
b) The distribution of the mean lifespan of 15 light bulbs is as described by the manufacturer. We can test it to see if it is drawn from the same population, but our sample doesn’t define the distribution. c) (10500-9000)/(1000/(sqrt(15))) = 5.81 There’s a very, very low probability that it’s greater than 10,500. d) Both distributions (I couldn’t get them to the same scale. The pop. distribution should be wider and larger.)
e) If the distribution were skewed, a small sample could not be described in terms of probabilities.
Problem 4.48
If you misjudged the N and in fact had 10 times as many samples, correcting it would lower the sigma, raising the Z score and lowering the p score.