Question

ANSWER ALL OF THE QUESTIONS WITH R CODE (SHOW ALL THE STEPS FOR THE R CODE!) THANK YOU!!

R problems For questions 1 and 2, load in and attach the Getting to Know You Survey for Spring 2018 using the following code: NoU <-read. csr(file = "http://users, stat . umn .edu/~kinne 174/Getting2NoUS 18 .csv", header = TRUE) attach(NoU) Assume that this survey is a random sample drawn from the population of interest which is all University of Minnesota students. 1. We are interested in determining the minimum sample size needed for a new survey to find a confidence interval for the true proportion p of students who are in favor of legalizing marijuana. We want a margin of error less than 3% and will use a 94% confidence interval. (a) We will need an estimate of p so we will use the sample proportion from our current survey. Calculate p using the following code table(legalizing.marijuana) margin.table(table(legalizing.marijuana))

Answer 1

Hi. There is no data attached with this question. So I will give you the code explaining what to do, but I can't run it and show you the results. If you have any doubts, please ask in the coments below.

I am answering Q2 as that is only complete question.

2.

a)

hist (height) ## This will create histogram of the data.

qqnorm(height) # This will create a normal Q-Q plot of the data

qqline(height) ## adds a theoritical line in the plot.

Histogram of height 50 60 70 80 90 height

Normal Q-Q Plot 3 210 2 Theoretical Quantiles

0 100 200300 400 Index

Shape: The data is approximately bell shaped curve. So normal assumption should be resonable.

Spread: The data is between 50 to 88 and with std deviation 4.44245.

Centre: The data is centered around 65~ 70. The mean of the data is 67.7.

Normality: The data can be considered as approximately normal as the q-q plot is a straight line and alomst coincide with the theoritical line.

EXPLANATION: If the data has histogram of above shape, then you can use t-test. Why?

Because the t-test confidence interval assumes the normality of the graph. So, if one is to use t-test, the data must be approximately normal. As $t = \frac{\sqrt{n}(\bar{x}-\mu)}{s}$ follows t(n-1) when X is normal.

So before applying the t-test, we test for normality by plotting the points or lokking at q-q plot.

b. Here we are trying to calculate the CI with t.test function. This function assumes normality of the data.

t.test (x = height, conf.level = .97)

here x= vector which contains height data and Conf level is desired level of confidence needed. You'll get the CI as result.

You can aslo try t.test (x = height, conf.level = .97)$conf.int to directly get CI

One Sample t-test

data: height
t = 332.84, df = 476, p-value < 2.2e-16
alternative hypothesis: true mean is not equal to 0
97 percent confidence interval:
67.25886 68.14433
sample estimates:
mean of x
67.70159

The 97% CI is 67.25886 68.14433

c.

for mean and std. dev, we have inbuild function

Use

mean(height) = 67.7

sd(height) = 4.44

d.

As for confidence interval, we need the two sided probability. So we break 3% into 1.5% and 1.5%

So for critical t value, we use

qt(0.985, df)

0.985 because of that 1.5% only in one sided tail

df = no. of observation -1 = n-1

e.

We have the formula for 97% CI for $\mu$

97% CI of $\mu$ is   $\bar{X} \mp t_{\alpha/2 , n-1}s.d. /\sqrt{n}$

The interpretaion is: By observing the data and under normality assumption, the probability that mean lies outside the given 97% interval is less than or equal to 3%.

So we have about 97% probability of getting our mean in that interval.

Thanks

Please give me thumbs up if you like, comment if you have any doubts.