6.6, 6.12, 6.20, 6.28, 6.44, 6.48

6.6: 2010 Healthcare Law

  1. False. We are 95% confident that the population mean exists between 43% and 49%. Our sample is 46% with no margin of error.

  2. True.

  3. True.

  4. False. As confidence goes down, margin of error goes down as well.

6.12: Legalization of marijuana, Part I

  1. 48% is a sample statistic since it represents the opinion of 48% of the sample drawn from the population.

  2. The 95% confidence interval is (45, 51). This means we are 95% confident that the population mean for the US approval rating for the legal use of marijuana exists between 45% and 51%.

  3. The critic’s concerns would be justified if we had a small sample, but our sample is large enough to negate these concerns.

  4. This statement is incorrect. The majority is barely covered in our confidence interval.

6.20: Legalize Marijuana, Part II

If we want to limit the margin of error of a 95% confidence interval to 2%, we will need to survey 2397 Americans.

6.28: Sleep deprivation, CA vs. OR, Part I

We are 95% confident that the population mean for CA exists between 7.5% and 8.5%. We are 95% confident that the population mean for OR exists between 8% and 9.6%.

These are proportions of residents in each state who reported insufficient rest or sleep during each of the preceding 30 days. Our confidence intervals for each state represent the interval in which the population mean can be found. We were only provided the sample mean.

6.44: Barking deer

  1. \[ H_o: \mu_1 = \mu_2 = \mu_3 = \mu_4 = \mu_5 \\ H_a: \mu_1 \neq \mu_2 \neq \mu_3 \neq \mu_4 \neq \mu_5 \]

  2. We can use a chi square test to answer this research question.

  3. Conditions: Sample observations are independent The samples are independent from each other. Sample size is large Our sample size is 426, so this condition is met. Population distribution is not strongly skewed.
    The population is fairly normal. Even if it wasn’t, we have enough samples to handle a skew.

  4. Our chi squared value is 284. Therefore, the data provide enough evidence to reject the null hypothesis in favor of the alternative.

6.48: Coffee and Depression

  1. Two-way chi squared test.

  2. H_o: The proportion of women who are depressed is the same across all levels of coffee consumption.
    H_a: The proportion of women who are depressed is different across varying levels of coffee consumption

  3. The proportion of women who suffer from depression is 0.05. The proportion of women who do not suffer from depression is 0.95.

  4. The expected value of the highlighted cell is 340. The contribution of this cell to the test statistic is 3.2.

  5. With a chi squared value of 20.93 and 4 degrees of freedom, our p-value is less than 0.005.

  6. The data provide enough evidence to conclude that there is a statistically significant difference in depression proportions of women across the varying levels of caffiene consumption. Therefore, we reject the null hypothesis.

  7. I agree with the author’s statement that it’s too early to recommend that women load up on extra coffee. Results from an experiment would likely produce results that we could use to suggest women drink coffee - but an observation study alone like this one isn’t enough.

1.96 * sqrt(0.48*(1-0.48)/1259)
## [1] 0.02759723
(0.48 - 0.028) * 100
## [1] 45.2
(0.48 + 0.028) * 100
## [1] 50.8
MJ.p <- .48
SE <- .02/1.96

(0.48 * (1-0.48))/((.02/1.96)^2)
## [1] 2397.158
#CA
1.96 * (sqrt((0.08)*(1-0.08)/11545))
## [1] 0.004948778
0.08 - 0.0049
## [1] 0.0751
0.08 + 0.0049
## [1] 0.0849
#OR
1.96 * (sqrt((0.088)*(1-0.088)/4691))
## [1] 0.008107036
0.088 - 0.0081
## [1] 0.0799
0.088 + 0.0081
## [1] 0.0961
(4-20.448)^2/20.448 + (16-62.622)^2/62.622 + (61-168.696)^2/168.696 +(345-174.234)^2/174.234
## [1] 284.0609
2607/50739
## [1] 0.05138059
48132/50739
## [1] 0.9486194
(2607/50739) * 6617
## [1] 339.9854
(373-((2607/50739) * 6617))^2/((2607/50739) * 6617)
## [1] 3.205914