Question

a poll of n voters is to be taken in an attempt to predict the outcome of a by - election in a certain riding. specifically, you are interested in the proportion of voters that will vote for a certain candidate, candidate a.

n = 403 voters have been randomly chosen, each has indicated what candidate they will vote for. you are to count the number, out of 403, who say they will vote for candidate a. the count is measured by the random variable x. you find x = 275.

(a) find a 90% confidence interval for p, the proportion of all voters who will vote for candidate a. use the z distribution to create your confidence interval. use at least four decimal points for your lower and upper bounds. at avoid rounding errors you should use r - studio and not tables.

lower bound =

upper bound =

(b) interpret the meaning of the interval you found in part (a).

the proportion of that will vote for candidate a is % and %. (enter your answer to at least two decimals.)

(c) find a 90% con e proportion of all voters who will vote for candidate a, by bootstrapping 1000 samples. use the seed 2568 to ensure that r - stud same
andom\ samples as this question will expect.

you can do this by including the code, you can copy it into your r - studio to bootstrap your samples.

rngkind(sample.kind =
ejection\)
set.seed(2568)
b = 1000
p = mean(resample(c(rep(1,275), rep(0,403 - 275)), 403))

find a 90% confidence interval. use at least four decimal points for your lower and upper bounds.

lower bound =

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}$ is calculated as $\hat{p}=\frac{X}{n}$, where $X = 275$ and $n=403$. So, $\hat{p}=\frac{275}{403}\approx0.6824$.

Step2: Find critical value

For a 90% confidence interval, the significance level $\alpha=1 - 0.90=0.10$, and $\alpha/2=0.05$. The critical value $z_{\alpha/2}$ from the standard normal distribution is $z_{0.05}\approx1.645$.

Step3: Calculate standard error

The standard error $SE=\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substituting $\hat{p}=0.6824$ and $n = 403$, we get $SE=\sqrt{\frac{0.6824\times(1 - 0.6824)}{403}}\approx\sqrt{\frac{0.6824\times0.3176}{403}}\approx\sqrt{\frac{0.2177}{403}}\approx0.0233$.

Step4: Calculate lower and upper bounds

The lower bound $LB=\hat{p}-z_{\alpha/2}\times SE$. Substituting the values, $LB = 0.6824-1.645\times0.0233=0.6824 - 0.0383=0.6441$.
The upper bound $UB=\hat{p}+z_{\alpha/2}\times SE$. Substituting the values, $UB=0.6824 + 1.645\times0.0233=0.6824+0.0383 = 0.7207$.

for part (b):
A confidence interval gives a range of values within which the True population parameter (in this case, the proportion of all voters who will vote for Candidate A) is likely to lie. A 90% confidence interval means that if we were to take many samples and construct confidence intervals in the same way, about 90% of those intervals would contain the True population proportion.

Answer:

Lower Bound = $0.6441$
Upper Bound = $0.7207$