Hypothesis Tests and Confidence Intervals for one and two proportions. 3. The na

Question

Hypothesis Tests and Confidence Intervals for one and two proportions. 3. The national average for the percentage of high school graduates taking the SAT is 49% A random sample of 300 high school graduating seniors were polled across a particular tristate area, and t was found that 195 had taken the SAT At alpha-0.05 level of significance, does the proportion of high school graduates who take the SAT in this area agree with the national average? Use a 95% confidence interval to estimate the true proportion take the SAT for this area. of high school graduates who b. c Find n if you want E to be 0.04. In a random sample of 200 men, 130 said they used seat belts. In a random sample of 300 women, 63 said they used seat belts. a. Test the claim that men are more safety conscious than women, at alpha 0.05 level of b. Find a 95% conf dence interval for the true deference in the proportions of the two groups. n1+m2 Tt2

Explanation / Answer

Q3.

PART A.

Given that,

possibile chances (x)=195

sample size(n)=300

success rate ( p )= x/n = 0.65

success probability,( po )=0.49

failure probability,( qo) = 0.51

null, Ho:p=0.49  

alternate, H1: p!=0.49

level of significance, = 0.05

from standard normal table, two tailed z /2 =1.96

since our test is two-tailed

reject Ho, if zo < -1.96 OR if zo > 1.96

we use test statistic z proportion = p-po/sqrt(poqo/n)

zo=0.65-0.49/(sqrt(0.2499)/300)

zo =5.5437

| zo | =5.5437

critical value

the value of |z | at los 0.05% is 1.96

we got |zo| =5.544 & | z | =1.96

make decision

hence value of | zo | > | z | and here we reject Ho

p-value: two tailed ( double the one tail ) - Ha : ( p != 5.54367 ) = 0

hence value of p0.05 > 0,here we reject Ho

ANSWERS

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null, Ho:p=0.49

alternate, H1: p!=0.49

test statistic: 5.5437

critical value: -1.96 , 1.96

decision: reject Ho

p-value: 0

they don't agree with the national average

PART B.

CI = confidence interval

confidence interval = [ 0.65 ± 1.96 * Sqrt ( (0.65*0.35) /300) ) ]

= [0.65 - 1.96 * Sqrt ( (0.65*0.35) /300) , 0.65 + 1.96 * Sqrt ( (0.65*0.35) /300) ]

= [0.596 , 0.704]

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interpretations:

1. We are 95% sure that the interval [ 0.596 , 0.704] contains the true population proportion

2. If a large number of samples are collected, and a confidence interval is created

for each sample, 95% of these intervals will contains the true population proportion

PART C.

Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)

Z a/2 at 0.05 is = 1.96

Sample Proportion = 0.65

ME = 0.04

n = ( 1.96 / 0.04 )^2 * 0.65*0.35

= 546.2275 ~ 547

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