Test the claims below. Identify the null hypothesis, alternative hypothesis, tes

Question

Test the claims below. Identify the null hypothesis, alternative hypothesis, test statistics, p-value or critical value(s), conclusion about the null hypothesis, and state a final conclusion that addresses the original claim. a) In an election, 208 out of 611 voters surveyed said they voted for the winning candidate. Use a 0.01 significance level to test the claim that among all voters, the percentage that believe they voted for the winning candidate is equal to 43%, which was the actual percentage of votes for the winner. What do your results indicate about voter perception here? b) A local school district claims that there is no difference between the mean performance of boys and girls at the 8th grade level. A random sample of scores of 10 students of each gender is listed below. Use a 0.05 significance level to test their claim. girls 90 80 70 75 87 92 86 61 94 100 boys 70 75 96 92 85 72 63 95 68 98 c) There is a new chief of police in the town of Low Gulch. Before his arrival, the town experienced an arrest rate of 25% for robberies. The new chief has compiled records showing that among the 30 recent robberies, the arrest rate is 30%, and claims that his rate is greater than the 25% of the past. Is there sufficient evidence to support his claim at the 0.05 significance level?

Explanation / Answer

a.
Given that,
possibile chances (x)=208
sample size(n)=611
success rate ( p )= x/n = 0.3404
success probability,( po )=0.43
failure probability,( qo) = 0.57
null, Ho:p=0.43
alternate, H1: p!=0.43
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.58
since our test is two-tailed
reject Ho, if zo < -2.58 OR if zo > 2.58
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.34043-0.43/(sqrt(0.2451)/611)
zo =-4.4723
| zo | =4.4723
critical value
the value of |z | at los 0.01% is 2.58
we got |zo| =4.472 & | z | =2.58
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -4.47232 ) = 0.00001
hence value of p0.01 > 0,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.43
alternate, H1: p!=0.43
test statistic: -4.4723
critical value: -2.58 , 2.58
decision: reject Ho
p-value: 0.00001

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