Chapter 6 - Inference for Categorical Data Practice: 6.5, 6.27, 6.43 Graded: 6.6, 6.28, 6.44

*6.6 2010 Healthcare Law. On June 28, 2012 the U.S. Supreme Court upheld the much debated 2010 healthcare law, declaring it constitutional. A Gallup poll released the day after this decision indicates that 46% of 1,012 Americans agree with this decision. At a 95% confidence level, this sample has a 3% margin of error. Based on this information, determine if the following statements are true or false, and explain your reasoning.

- We are 95% confident that between 43% and 49% of Americans in this sample support the decision of the U.S. Supreme Court on the 2010 healthcare law.

False. A confidence interval is constructed to estimate the population proportion, we know that 46% of our sample support this decision.

- We are 95% confident that between 43% and 49% of Americans support the decision of the U.S. Supreme Court on the 2010 healthcare law.

True. Since we have a 3% margin of error at the 95% confidence interval we can assume with 95% confidence that the 46% ± 3% support the decision.

- If we considered many random samples of 1,012 Americans, and we calculated the sample proportions of those who support the decision of the U.S. Supreme Court, 95% of those sample proportions will be between 43% and 49%.

False. 95% of samples will include the true population proportion of Americans who support the decision of the Supreme Court.

- The margin of error at a 90% confidence level would be higher than 3%.

False. z value of 90% confidence interval is lower that 95%, so our margin or error will be lower.

*6.28 Sleep deprivation, CA vs. OR, Part I. According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insu

```
SE <- sqrt( ((0.08 * (1 - 0.08)) / 11545) + ((0.088 * (1 - 0.088)) / 4691))
#z for 95% CI is 1.96
ME = 1.96*SE
round(0.088-0.08-ME,4)
```

`## [1] -0.0015`

`round(0.088-0.08+ME,4)`

`## [1] 0.0175`

Since our confidence interval includes 0 - we cannot reject the null hypothesis and we conlude the sleep deprivation in California and Oregon is not significantly different.

*6.44 Barking deer. Microhabitat factors associated with forage and bed sites of barking deer in Hainan Island, China were examined from 2001 to 2002. In this region woods make up 4.8% of the land, cultivated grass plot makes up 14.7% and deciduous forests makes up 39.6%. Of the 426 sites where the deer forage, 4 were categorized as woods, 16 as cultivated grassplot, and 61 as deciduous forests. The table below summarizes these data.62 Woods Cultivated grassplot Deciduous forests Other Total 4 16 67 345 426

- Write the hypotheses for testing if barking deer prefer to forage in certain habitats over others.

H0: Barking Deer are proportionally distributed over the various types of land H1: Barking Deer are more likely to forage in particular type of land

- What type of test can we use to answer this research question?

We can use a chi-squared goodness of fit test.

- Check if the assumptions and conditions required for this test are satisfied.

The assumptions and conditions are satisfied: - The observations are independent, we assume there is no dependence between the cases of deer distribution we are considering - We have at least 5 expected cases for each scenario. Woods have 0.048*426 = 20.5 cases.

- Do these data provide convincing evidence that barking deer prefer to forage in certain habitats over others? Conduct an appro- priate hypothesis test to answer this research question.

```
df <- 4-1
#Proporiton of "other"
1-0.048-0.147-0.396
```

`## [1] 0.409`

```
chi<-((4-0.048*426)^2)/(0.048*426)+((16-0.147*426)^2)/(0.147*426)+((67-0.396*426)^2)/(0.396*426)+((345-0.409*426)^2)/(0.409*426)
p_Val <- pchisq(chi, 3, lower.tail = FALSE)
p_Val
```

`## [1] 1.144396e-59`

Since the p-value is less than 5%, we reject our null hypothesis. The data provides convincing evidence that the deer are not proportionally distributed over various types of land.