- False. We are 100% confident that exactly 46% of the sample support the decision.
- True. This is the accurate use of a confidence interval.
- False. The confidence interval describes the population proportion, not other confidence intervals.
- False. If we are less confident in our answer, we can select a smaller range of values. In order to be more confident, we must select more values.
- This is the sample statistics. 48% represents the amount of people in the sample that are in favor of legalization.
n <- 1259
p <- 0.48
n*p
## [1] 604.32
n*(1-p)
## [1] 654.68
ME <- sqrt(p*(1-p)/n)*1.96
p - ME
## [1] 0.4524028
p + ME
## [1] 0.5075972
- The sample is less than 10% of the population. We can assume an independence. np and n(1-p) are greater than 10. Thus, the confidence for inference are met.
It can be stated with 95% confidence that the proportion of Americans that support marijuana legalization is between 45.24% and 50.76%
- Yes. The conditions for inference have been met in part b. Thus normality can be assumed.
- Yes. Percentages including above the 50% mark are included in the confidence interval. This is a fair assessment.
\[1.96\times\sqrt{\frac{0.48(1-0.48)}{n}}=0.02, n\approx2398\] 2398 Americans would need to be survered.
pc <- 0.08
nc <- 11545
po <- 0.088
no <- 4691
CV <- 1.96
pc*nc
## [1] 923.6
(1-pc)*nc
## [1] 10621.4
po*no
## [1] 412.808
(1-po)*no
## [1] 4278.192
ME <- sqrt(pc*(1-pc)/nc + po*(1-po)/no)
pd <- po-pc
pd + ME
## [1] 0.01284598
pd - ME
## [1] 0.003154016
The sample is less than 10% of the population. We can assume an independence. np and n(1-p) are greater than 10. Thus, the conditions for inference are met.
It can me stated with with 95% confidence that the difference in percentage of Oregan residents vs California residents who are sleep deprived is between 0.3% and 1%.
- \(H_0:\) Barking deer have no preference in foraging location \(H_a:\) Barking deer disproportionately favor certain foraging locations over others.
- We should use a chi-squred test to determine whether or not to reject the null hypothesis.
.048*426
## [1] 20.448
- The sampling method is a simple random sample. The variables are categorical. The expected value of observations in each level is at least 5. Thus, the conditions for inference are met.
found <- c(4, 16, 61, 345)
expected <- c(20.448, 62.622, 168.696, 174.234)
sum((found - expected)^2 / expected)
## [1] 284.0609
- With a \(X^2\) of approximately 284 and 3 degrees of freedom, the p value is less than 0.001.
With a p-value so small we reject the null hypothesis. It can be stated with strong confidence that barking deer prefer to forage in certain habitats.
- A chi-squared test is appropriate.
2607/50739
## [1] 0.05138059
48132/50739
## [1] 0.9486194
- \(H_0:\) There is no relationship between amount of coffee consumed and depression in women \(H_a:\) There is a disproportionate rate of depression in women who drink certain amounts of coffee.
expected <- 6617*2607/50739
observed <- 373
(observed-expected)^2 / expected
## [1] 3.205914
- Approximately 5.14% of women in the sample suffer from depression while approximately 94.86% do not.
- Expected value: 339.9854 A total of appoximately 3.2 is added to the test statistic.
- With a \(X^2\) of 20.93 and 4 degrees of freedom the p-value is less than 0.001
- With a p-value less than 0.001 we reject the null hypothesis. There is strong evidence to suggest that there is a relationship between coffee consumption and depression in women.
- Yes I do. The researchers have found a correlation between the variables but they have no found causation. That is, there is no evidence to suggest that drinking more coffee reduces depression. There could be an underlying or confounding variables. For example, women who lead fulfilling and busy lives are both less likely to have depression and more likely to need coffee.