A polling agency says that over 75% of people in the United States prepare and f

Question

A polling agency says that over 75% of people in the United States prepare and file their income taxes before April 15th. In a random survey 1000 people in the United States, 790 said that they prepare and file their income taxes before April 15th. Let p be the true proportion of people in the United States prepare and file their income taxes before April 15th. To test the agency's claim, which of the following is correct? H_0: p = 0.79 versus H_a: p > 0.79 H_0: p = 0.79 versus H_a p not equal to 0.79 H_0: p = 0.75 versus H_a: p not equal to 0.75 H_0: p = 0.75 versus H_a: p > 0.75 H_0: p = 0.75 versus H_a: p > 0.79 The standardized test statistics is about (a) -3.11 (b) -2.92 (c) 2.92. (d) 3.11 (e) 4.11

Explanation / Answer

Given that,
possibile chances (x)=790
sample size(n)=1000
success rate ( p )= x/n = 0.79
success probability,( po )=0.75
failure probability,( qo) = 0.25
null, Ho:p=0.75
alternate, H1: p>0.75
level of significance, = 0.05
from standard normal table,right tailed z /2 =1.64
since our test is right-tailed
reject Ho, if zo > 1.64
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.79-0.75/(sqrt(0.1875)/1000)
zo =2.9212
| zo | =2.9212
critical value
the value of |z | at los 0.05% is 1.64
we got |zo| =2.921 & | z | =1.64
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail - Ha : ( p > 2.92119 ) = 0.00174
hence value of p0.05 > 0.00174,here we reject Ho

ANSWERS
---------------
24.

null, Ho:p=0.75
alternate, H1: p>0.75
25.

test statistic: 2.9212
critical value: 1.64
decision: reject Ho
p-value: 0.00174

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