Question

can you explain 5d and 5e, thank you, I will leave a good rating!

5d. In a comparative study of two new drugs, A and B, 300 patients were treated with drug A, and 275 patients were treated with drug B. (The two treatment groups were randomly and independently chosen.) It was found that 222 patients were cured using drug A and 217 patients were cured using drug B. Let p1 be the proportion of the population of all patients who are cured using drug A, and let p2 be the proportion of the population of all patients who are cured using drug B. Find a 90% confidence interval for −p1p2. Then complete the table below.

Carry your intermediate computations to at least three decimal places. Round your responses to at least three decimal places.

What is the lower limit of the 90% confidence interval? What is the upper limit of the 90% confidence interval?

5e. One personality test available on the World Wide Web has a subsection designed to assess the "honesty" of the test-taker. After taking the test and seeing your score for this subsection, you're interested in the mean score, μ, among the general population on this subsection. The website reports that μ is 148, but you believe that μ differs from 148. You decide to do a statistical test. You choose a random sample of people and have them take the personality test. You find that their mean score on the subsection is 143 and that the standard deviation of their scores is 22.

Based on this information, answer the questions below.

What are the null hypothesis (H0) and the alternative hypothesis (H1) that should be used for the test? H0: μ is ?less than less than or equal to greater than greater than or equal to not equal to equal to ?22,148 or 143 H1: μ is ?less than less than or equal to greater than greater than or equal to not equal toe qual to ?22, 148 or 143 In the context of this test, what is a Type I error? A Type I error is ?rejecting failing to reject the hypothesis that μ is ?less than less than or equal to greater than greater than or equal to not equal to equal to ?22148143 when, in fact, μ is ?less than less than or equal to greater than greater than or equal to not equal to equal to ?22148143. Suppose that you decide not to reject the null hypothesis. What sort of error might you be making? ?Type l or Type II

Answer 1

5d.

TRADITIONAL METHOD
given that,
sample one, x1 =222, n1 =300, p1= x1/n1=0.74
sample two, x2 =217, n2 =275, p2= x2/n2=0.789
I.
standard error = sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where
p1, p2 = proportion of both sample observation
n1, n2 = sample size
standard error = sqrt( (0.74*0.26/300) +(0.789 * 0.211/275))
=0.035
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.64
margin of error = 1.64 * 0.035
=0.058
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.74-0.789) ±0.058]
= [ -0.107 , 0.009]
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DIRECT METHOD
given that,
sample one, x1 =222, n1 =300, p1= x1/n1=0.74
sample two, x2 =217, n2 =275, p2= x2/n2=0.789
CI = (p1-p2) ± sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where,
p1, p2 = proportion of both sample observation
n1,n2 = size of both group
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ (0.74-0.789) ± 1.64 * 0.035]
= [ -0.107 , 0.009 ]
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interpretations:
1) we are 90% sure that the interval [ -0.107 , 0.009] contains the difference between
true population proportion P1-P2
2) if a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the difference between
true population mean P1-P2
5e.
Given that,
population mean(u)=148
sample mean, x =143
standard deviation, s =22
number (n)=20
null, Ho: μ=148
alternate, H1: μ!=148
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.093
since our test is two-tailed
reject Ho, if to < -2.093 OR if to > 2.093
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =143-148/(22/sqrt(20))
to =-1.0164
| to | =1.0164
critical value
the value of |t α| with n-1 = 19 d.f is 2.093
we got |to| =1.0164 & | t α | =2.093
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -1.0164 ) = 0.3222
hence value of p0.05 < 0.3222,here we do not reject Ho
ANSWERS
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null, Ho: μ=148
alternate, H1: μ!=148
test statistic: -1.0164
critical value: -2.093 , 2.093
decision: do not reject Ho
p-value: 0.3222
we do not have enough evidence to support the claim that mean is 148.
In this context,
type 2 error is possible because it fails to reject the null hypothesis.