Carry the analysis when 25 out of the 324 cases were false positives with full h

Question

Carry the analysis when 25 out of the 324 cases were false positives with full hypothesis testing. PLEASE USE 25 IN YOUR ANSWER INSTEAD OF 23

Carry out a full hypothesis test which includes

1. Ho and Ha.

2. State assumptions. Comments about the validity of assumptions. For example, if normality is not mention in the problem, you should mention that there is no enough information to check this assumption.

3. Find the value of test-statistic.

4. Distribution of test statistic when H0 is true (normal or t-distribution. Remember to write df for t-distribution).

5. Always find p-value

6. Conclusion and answer the question

10.1.8 In trials of a medical screening test for a particular illness, 23 cases out of 324 positive results turned out to be "false-positive" results. The screening test is acceptable as long as p, the probability of a positive result being incorrect, is no larger than 10%. Calculate a p-value for the hypotheses Ho p20.1 versus HA:p

Explanation / Answer

PART A.
Given that,
possibile chances (x)=25
sample size(n)=324
success rate ( p )= x/n = 0.0772
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,left tailed z /2 =1.64
since our test is left-tailed
reject Ho, if zo < -1.64
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.07716-0.1/(sqrt(0.09)/324)
zo =-1.3704
| zo | =1.3704
critical value
the value of |z | at los 0.05% is 1.64
we got |zo| =1.37 & | z | =1.64
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: left tail - Ha : ( p < -1.37037 ) = 0.08529
hence value of p0.05 < 0.08529,here we do not reject Ho
ANSWERS
---------------
null, Ho:p>=0.1
alternate, H1: p<0.1
test statistic: -1.3704
critical value: -1.64
decision: do not reject Ho
p-value: 0.08529

PART B.
given that,
success rate ( p )= x/n = 0.0772
CI = confidence interval
confidence interval = [ 0.0772 ± 2.576 * Sqrt ( (0.0772*0.9228) /324) ) ]
= [0.0772 - 2.576 * Sqrt ( (0.0772*0.9228) /324) , 0.0772 + 2.576 * Sqrt ( (0.0772*0.9228) /324) ]
= [0.039 , 0.1153]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 99% sure that the interval [ 0.039 , 0.1153] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population proportion

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