I have calculated everything, I just need help finding the P-value please! A 201

Question

I have calculated everything, I just need help finding the P-value please!

A 2010 survey by a reputable automotive website found that 61% of vehicle owners avoided automotive maintenance and repairs. Suppose a company would like to perform a hypothesis test to challenge this finding. From a random sample of 160 vehicle owners, it was found that 114 avoid maintenance repairs. Using alpha = 0.10, answer parts a and b below. What conclusions can be drawn concerning the proportion of vehicle owners who avoid automotive maintenance and repairs? Determine the null and alternative hypotheses. Choose the correct answer below. H_0:p > 0.61 H_1:p lessthanorequalto 0.61 H_0:p greaterthanorequalto 0.61 H_1:p 0.61 Determine the critical value(s) of the test statistic. z_alpha = -1.645, 1.645 (Use a comma to separate answers as needed. Round to three decimal places as needed.) Calculate the test statistic. z_p = 2.67 (Round to two decimal places as needed.) Determine the conclusion. Choose the correct answer below. Reject H_0. There is not sufficient evidence to conclude that the true proportion of owners who avoid maintenance and repairs is different from what the website found. Do not reject H_0. There is sufficient evidence to conclude that the true proportion of owners who avoid maintenance and repairs is different from what the website found. Do not reject H_0. There is not sufficient evidence to conclude that the true proportion of owners who avoid maintenance and repairs is different from what the website found. Reject H_0. There is sufficient evidence to conclude that the true proportion of owners who avoid maintenance and repairs is different from what the website found. Determine the p-value for this test. p-value = _____ (Round to three decimal places as needed.)

Explanation / Answer

Given that,
possibile chances (x)=114
sample size(n)=160
success rate ( p )= x/n = 0.71
success probability,( po )=0.61
failure probability,( qo) = 0.39
null, Ho:p=0.61
alternate, H1: p!=0.61
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.64
since our test is two-tailed
reject Ho, if zo < -1.64 OR if zo > 1.64
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.7125-0.61/(sqrt(0.2379)/160)
zo =2.66
| zo | =2.66
critical value
the value of |z | at los 0.1% is 1.64
we got |zo| =2.658 & | z | =1.64
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 2.65819 ) = 0.00786
hence value of p0.1 > 0.0079,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.61
alternate, H1: p!=0.61
test statistic: 2.66
critical value: -1.64 , 1.64
decision: reject Ho
p-value: 0.00786

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