4 It has been reported that 10% of adults in the United States suffer from major

Question

4 It has been reported that 10% of adults in the United States suffer from major depressive illness. diabetes have a higher percentage of depression. To test this belief, she reviews health records and conducts interviews with 350 randomly selected diabetes patients and finds that 45 have major depressive illness. Does the sample data support the official's beliefs? Test at the 0.05 A health official believes that people with significance level. a. Describe the population parameter of interest. b. State the null (Ho)& alternative (H) hypotheses. He: Ha: c. Check the assumptions, identify the probability distribution for this test, and state the significance level d. What is the point estimate, p', for the percent of adults in Tennessee suffering from occasional insomnia? e. Calculate the value of the test statistic, * f. state the p-value, g. and define the critical region. h. Based on this information, state the decision i. and the conclusion. j. If this had been a two-sided alternative hypothesis, would our results be different? If so, how?

Explanation / Answer

4.
a. population parameter of interest
Given that,
possibile chances (x)=45
sample size(n)=350
success rate ( p )= x/n = 0.1286
success probability,( po )=0.1
failure probability,( qo) = 0.9
null, Ho:p=0.1  
alternate, H1: p>0.1
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=0.12857-0.1/(sqrt(0.09)/350)
zo =1.7817
| zo | =1.7817
critical value
the value of |z | at los 0.05% is 1.64
we got |zo| =1.782 & | z | =1.64
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail - Ha : ( p > 1.78174 ) = 0.0374
hence value of p0.05 > 0.0374,here we reject Ho
ANSWERS
---------------
b.
null, Ho:p=0.1
alternate, H1: p>0.1
c.
Z test for proportion at significance level is 0.05
d.
possibile chances (x)=45
sample size(n)=350
success rate ( p )= x/n = 0.1286
e.
test statistic: 1.7817
f.
p-value: 0.0374
g.
critical value: 1.64
h.
decision: reject Ho
i.
we have enough evidence to support the health official believes that a people with diabets have higher percentage of depression
j.
Given that,
possibile chances (x)=45
sample size(n)=350
success rate ( p )= x/n = 0.1286
success probability,( po )=0.1
failure probability,( qo) = 0.9
null, Ho:p=0.1  
alternate, H1: p!=0.1
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=0.12857-0.1/(sqrt(0.09)/350)
zo =1.7817
| zo | =1.7817
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =1.782 & | z | =1.96
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 1.78174 ) = 0.07479
hence value of p0.05 < 0.0748,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.1
alternate, H1: p!=0.1
test statistic: 1.7817
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.07479

if we take two tailed hypothesis compare with single side it could changed the decision and result also

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