Problem 6.6
a) Our confidence interval is for the statistical inference and not for the sample itself.
b) This is true. To find the confidence interval, we add and subtract the margin of error.
c) This is not knowable from getting only one sample. We are 95% sure the true mean is in our interval. We do not have knowledge of other samples.
d) False. The smaller the percentage of certitude, the smaller the margin of error.
Problem 6.12
a) 48% is a simple statistic. No inference was made.
b) The 95% confidence interval ranges from 45.24% to 50.76%. We believe, with 95% confidence, that the percentage of Americans who believe marijuana should be legal is in that range.
c) This is a reasonable assumption. There are enough samples to expect the law of large numbers means our sample should be normal. The survey randomly chose independent participants.
d) This is not a reasonable assumption. Only a small part of our confidence interval is over 50%. Our sample mean is under 50%. It seems more than likely that the headline is false.
Problem 6.20
We need to sample 2398 people. .02= 1.96* sqrt((.48)(.52)/n) n = 2398
Problem 6.28
The 95% confidence interval for the percentage of Oregonians who are expected to be sleep deprived in excess of the percentage of Californians who are is .15% to 1.75%. The calculation is .008 +/- sqrt ((.08)(.92)/11545 + (.088)(.912)/4691).
Problem 6.44
a) H0: The foraging choices of the barking deer in Hainan Island represent the avaliable proportions of habitat.
H1: The foraging choices of the barking deer in Hainan Island do not represent the avaliable proportions of habitat. b) A chi-squared test will allow us to test if the grouped data match the expected proportions of grouped data.
c) Independence seems to be fulfilled because the deer were free to choose where to forage. There are only 4 sites in the woods group. It should be grouped with another group to satisfy the conditions. “Other” and “deciduous forests” are the most appropriate groupings. It would seem woods would go well with deciduous forests, but better knowledge of Hainan Province would help make a better determintion.
d) Expecteds: wood,deciduous forest 189.14; cultivated grass plot 62.62; other 174.23
Chi square stat: ((71-189.14)2 / 189.14) + ((16-62.62)2 / 62.62) + ((345-174.23)2 / 174.23), which equals 275.8790775.
Compare this to a 11.345 critical value for a chi-square distribution with 3 degrees of freedom. Our null hypothesis. It seems clear that our deer prefer “other” microhabitat for foraging.
Problem 6.48
a) A chi-square test is the appropriate test for grouped data when we suspect a part of the population may have a proportion of a factor different from that of the greater population.
b) H0 : Drinking different amounts of coffee is not associated with depression incidence in women.
H1 : Drinking different amounts of coffee is associated with depression incidence in women.
c) 5.14% of the women in the study experienced depression.
d) 339.99 women were expected in the 2-6 cups per week, yes category. This contributes ((373-339.99)2)/339.99 = 3.2049769 to our chi-square statistic.
e) For a chi-squared distribution with 4 degrees of freedom, a p-value of .001 corresponds to 18.4668. At 20.93, our test result is further into the tail than that.
f) We reject the null hypothesis and find that there was an association between coffee use and depression among woman in our study.
g) Our researcher was correct. While there is an association, we can not rule out other factors that may have lead to this outcome. Perhaps heavy coffee drinkers share some other factors that lead to lower depression.