Rider University W 5 Proportions Questions Students must show all work in order to receive credit. Week 5 – More Proportions Chapter 21 goes into detail

Rider University W 5 Proportions Questions Students must show all work in order to receive credit.

Week 5 – More Proportions

Chapter 21 goes into detail about some of the nuances of hypothesis testing. Of particular importance are the following points:

P-values are conditional probabilities. They tell us the probability of getting a result at least as unusual as the observed statistics, given that the null hypothesis is true.
Generally speaking, we operate with a 95% confidence level, which would equate to an alpha of .05. There are occasions, though, where we want to be even more certain (i.e. 99% confidence interval, corresponding to an alpha of .01), or less certain. This alpha level is often called the significance level.
Confidence intervals and hypothesis tests (two-tailed) are very similar. Actually, if you want to perform a two-tailed hypothesis test, another approach is to create a confidence interval, and determine whether the observed statistic is within (don’t reject) or outside of (reject) that interval.
There is built in error to hypothesis testing. Two types of errors in particular are noted: Type 1 – when the null is true, but we reject it; and Type 2 – when the null is false, but we fail to reject it. There is tension between Type 1 and Type 2 errors, and the only way to reduce both types of errors is to collect more data.
Power tells us the probability that a test correctly rejects a false null, and the effect size is the distance between the null hypothesis value and the truth. Week 5 Discussion Board
a. Answer the following question, making sure to explain your thinking in detail.
Researchers contacted more than 25,000 Americans aged 24 years to see if they had finished
high school; 84.9% of the 12,460 males and 88.1% of the 12,678 females indicated that they had
high school diplomas.
1. Create a 95% confidence interval for the difference in graduation rates between males
and females.
2. Interpret your confidence interval.
3. Does this provide strong evidence that girls are more likely than boys to complete high
school? Explain.
b. Find one aspect of this week’s material that is relevant to college, career, or everyday life.
Provide some detail on how it could be important.
Week 5 Problem Set:
1.
Someone flipped a coin 250 times, and got heads 140 times. Let’s test a
hypothesis about the fairness of this coin.
a. Estimate the true proportion of heads. Use a 95% confidence interval.
b. Does your confidence interval provide evidence that the coin is unfair?
Explain.
c. What is the significance level of this test? Explain.
2.
The AAP examined parental influence on teenagers’ decisions to smoke. A group
of students who never smoked were questioned about their parents’ attitudes
toward smoking. These students were questioned again two years later to see if
they had started smoking. The researchers found that, among the 284 students
who indicated that their parents disapproved of kids smoking, 54 had become
established smokers. Among the 41 students who initially said their parents were
lenient about smoking, 11 became smokers.
a. Create a 95% confidence interval for the difference in proportion of
children who may smoke and have approving parents and those who may
smoke and have disapproving parents.
b. Interpret your interval in this context.
c. Explain what 95% confidence means.
3.
A study examined the impact of depression on a patient’s ability to survive
cardiac disease. Researchers identified 450 people with cardiac disease,
evaluated them for depression, and followed the group for 4 years. Of the 361
patients with no depression, 67 died. Of the 89 patients with minor or major
depression, 26 died. Among the patients who suffer from cardiac disease, are
depressed patients more likely to die than non-depressed ones?
a. Write appropriate hypotheses.
b. Test your hypothesis and state your conclusion.
c. Explain in this context what your P-value means.
d. If your conclusion is actually incorrect, which type of error did you
commit?
Week 5 – More Proportions
Chapter 21 goes into detail about some of the nuances of hypothesis testing. Of particular
importance are the following points:
– P-values are conditional probabilities. They tell us the probability of getting a result at
least as unusual as the observed statistics, given that the null hypothesis is true.
– Generally speaking, we operate with a 95% confidence level, which would equate to an
alpha of .05. There are occasions, though, where we want to be even more certain (i.e.
99% confidence interval, corresponding to an alpha of .01), or less certain. This alpha
level is often called the significance level.
– Confidence intervals and hypothesis tests (two-tailed) are very similar. Actually, if you
want to perform a two-tailed hypothesis test, another approach is to create a confidence
interval, and determine whether the observed statistic is within (don’t reject) or outside of
(reject) that interval.
– There is built in error to hypothesis testing. Two types of errors in particular are noted:
Type 1 – when the null is true, but we reject it; and Type 2 – when the null is false, but
we fail to reject it. There is tension between Type 1 and Type 2 errors, and the only way
to reduce both types of errors is to collect more data.
– Power tells us the probability that a test correctly rejects a false null, and the effect size is
the distance between the null hypothesis value and the truth.
Let’s do an example:
Looking back at assumptions and conditions (p. 446). We can assume independence, based on
randomization and the 10% condition (1302 is fewer than 10% of men). We can assume
sufficient sample size, since 39 and 1302-39 are both at least 10.
a. ??? , here, would be 39/1302 = .03. SE can be calculated with sqrt(??? ???/n), or
sqrt(.03*.97/1302) = .0047. To find a 98% confidence interval, we need to know how
many z-scores above and below the mean we are talking about. If I were to draw a
normal curve, there would be 1% on the left and 1% on the right. So, then, I will find the
z-score that is associated with the 1st percentile, using =NORMSINV(.01), and I get –
2.33. Therefore, we need to do ??? ± 2.33 SE, or .03 ± 2.33 * .0047, or .019 to .041. We
are 98% confident that between 1.9% and 4.1% of men measure success primarily from
their work.
b. Yes. The confidence interval indicates that there is strong evidence that fewer than 5% of
all men judge their success primarily by their work.
c. The significance level of this test is .01, since we are utilizing a lower-tail test.
Chapter 22 shows us a situation that is much more frequent – when we want to make
comparisons between two percentages. The challenge, here is that we need to consider the
standard deviation of both samples to gauge the standard deviation between them. The tool we
are going to use is described on pp. 504-506, and it included here for convenience:
More generally, to perform a two-proportion z-interval test, we check assumptions and
conditions, and then utilize the following:
When doing hypothesis testing, we need to utilize the concept of pooling, which involves
combining the counts to get an overall proportion. Please see the Teen Smoking example below,
and see the steps for a two-proportion z-test.
Let’s do a confidence interval, then a hypothesis test.
a. We have randomization and the 10% condition, and the groups are independent of one
another. In addition, there are 10 successes and failures in each group.
b. We need ??? women – ??? men, which is 535/1062 – 411/1012 = .098. Our standard error, using
the big-time formula above, will be sqrt(.504*.496/1062 + .406*.594/1012) = .022. We
need, then to do .098 plus/minus 1.96 * .022, which gives us .055 to .141.
c. We are 95% confident that the proportion of men and women who have Arthritis differs
by between 5.5% and 14.1%.
d. Yes, since the 95% confidence interval suggests all positive values.
a. This is an observational study.
b. The null hypothesis: the proportion of teens whose parents disapprove of smoking – the
proportion of teens whose parents are lenient of smoking is 0. Alternate hypothesis: the
proportion of teens whose parents disapprove of smoking – the proportion of teens whose
parents are lenient of smoking