Given:
p=0.46, n=1012, me=0.03
Given:
n=1259, p=0.48
statistic vs. paramater: statistic is a measurement for a sample and paramater is a measurement for a population. The 48% is for a given sample, then it’s a sample statistic.
n<-1259
p<-0.48
confidence<-0.95
se<-sqrt(p*(1-p)/n)
z<-qt(confidence+((1-confidence)/2),n-1)
me<-z*se
low<-p-me
up<-p+me
For the 95% confidence interval, it’s between (0.4523767, 0.5076233).
It is true because to have a normal model to be a good approximation, we need to validate: 1- the # of successes: np >= 10, np = 12590.48 = 604.32 2- the # of the failures: n(1-p) >= 10, n(1-p) = 1259(1-0.48) = 654.68
No. Based on the confidence interval calculated above in this question, we can state that it’s up to 50% and it doesn’t represent the majority.
Given:
CA: \(p_1=0.08\), \(n_1=11545\),
OR: \(p_2=0.088\), \(n_2=4691\),
confidence=0.95 => z=1.96
Margin of Error Formula: \(ME = z * sqrt{(p_1*(1-p_1)/n_1)+(p_2*(1-p_2)/n_2)}\)
p1<-0.08
n1<-11545
p2<-0.088
n2<-4691
me<-z*sqrt((p1*(1-p1)/n1)+(p2*(1-p2)/n2))
low<-(p1-p2)-me
up<-(p1-p2)+me
For the 95% confidence interval for the difference between the proportions of Californians and Oregonians who are sleep deprived, it’s between (-0.0175, 0.0015).
Given:
n=426
\(p_1=0.048\)
\(p_2=0.147\)
\(p_3=0.396\)
\(H_0\) - the population proportions are equal to the mentioned proportions:
\(p_1=0.048\), \(p_2=0.147\), \(p_3=0.396\) and \(p_4=1-0.048-0.147-0.396=0.409\)
\(H_1\) - the population proportions are not equal to the mentioned proportions:
at least one of the \(p_i\) has a different value.
we can use the chi-square test, since we have cases that can be classified into several groups.
Independence: we assume that the plots are independent of each other Sample size / distribution: expected values are all at least 5 expected cases.
Formulas:
\(E = n * p_i\),
\(chi^2 = (O - E)^2 / E\), where O = observed frequency and E = expected frequency
n<-426
p<-c(0.048, 0.147, 0.396, 0.409)
observed<- c(4, 16, 61, 345)
expected<- c(n*p[1], n*p[2], n*p[3], n*p[4])
chi_square <- (observed - expected )^ 2 / expected
chi<-sum(chi_square)
p_value <- 1 - pchisq(chi, df = length(observed)-1)
p_value = 0.
p_value < 0.001 & p_value < 0.05 => reject \(H_0\).
There’s enough evidence to support the claim that barking deer forage in certain habitats over others.