Question

a group of nonsmokers exposed to tobacco smoke (Summary Stats: n = 40, x = 60.58 ng/mL, and s = 138.08 ng/mL) and a group of nonsmokers not exposed to tobacco smoke (Summary Stats: n = 40, x = 16.35 ng/mL, and s = 62.53 ng/mL). Use a 0.05 significance level to test the claim that nonsmokers exposed to tobacco smoke have a higher mean cotinine level than nonsmokers not exposed to tobacco smoke. (Assume the two samples are independent simple random samples selected from a normally distributed population and with not equal standard deviations as typical of these types of problems.)

State the claim using words (i.e., “The claim is [insert the English words here]”.)

Write the null (H0) and alternative (H1) hypotheses using symbols.

Indicate whether H0 or H1 is the claim by writing “(claim)” next to whichever one it is.

State the tail of the hypothesis test: left-tailed/one-tailed, right-tailed/one-tailed, or two-tailed.

State the level of significance. (i.e., a = __) State the degrees of freedom (i.e., DF = __) or write “not applicable” if not relevant.

State all the requirements with respect to this type of problem.

Verify any requirement that should require a calculation. If none, write “not applicable”.

Identify which test statistic to use and write the associated formula using symbols.

Compute the value of the test statistic. If using StatCrunch, circle it. State the P-value.

Show and apply the P-value decision rules (with values substituted appropriately) that lead you to “reject H0” or “fail to reject H0”.

State the critical value(s).

Show and apply the critical value decision rules (with values substituted appropriately) that lead you “reject H0” or “fail to reject H0”.

Write the conclusion utilizing the “correct” words using Table 8-3, p. 366. Make sure you insert the relevant portions of the claim into the conclusion.

Construct the associated confidence interval (CI).

State the CI using the correct notation using the correct notation. State the margin of error (E = __).

State the point estimate using the correction notation.

Does the CI support the hypothesis test conclusion? Explain/interpret.

Express the Type I error in the context of the problem

Express the Type II error in the context of the problem

Answer 1

a.

Given that,
mean(x)=60.58
standard deviation , s.d1=138.08
number(n1)=40
y(mean)=16.35
standard deviation, s.d2 =62.53
number(n2)=40
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.685
since our test is right-tailed
reject Ho, if to > 1.685
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =60.58-16.35/sqrt((19066.0864/40)+(3910.0009/40))
to =1.8455
| to | =1.8455
critical value
the value of |t α| with min (n1-1, n2-1) i.e 39 d.f is 1.685
we got |to| = 1.84548 & | t α | = 1.685
make decision
hence value of | to | > | t α| and here we reject Ho
p-value:right tail - Ha : ( p > 1.8455 ) = 0.03629
hence value of p0.05 > 0.03629,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 1.8455
critical value: 1.685
decision: reject Ho
p-value: 0.03629
we have enough evidence to support the claim that nonsmokers exposed to tobacco smoke have a higher mean cotinine level than nonsmokers not exposed to tobacco smoke.
b.
TRADITIONAL METHOD
given that,
mean(x)=60.58
standard deviation , s.d1=138.08
number(n1)=40
y(mean)=16.35
standard deviation, s.d2 =62.53
number(n2)=40
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((19066.086/40)+(3910.001/40))
= 23.967
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.05
from standard normal table,right tailed and
value of |t α| with min (n1-1, n2-1) i.e 39 d.f is 1.685
margin of error = 1.685 * 23.967
= 40.384
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (60.58-16.35) ± 40.384 ]
= [3.846 , 84.614]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=60.58
standard deviation , s.d1=138.08
sample size, n1=40
y(mean)=16.35
standard deviation, s.d2 =62.53
sample size,n2 =40
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 60.58-16.35) ± t a/2 * sqrt((19066.086/40)+(3910.001/40)]
= [ (44.23) ± t a/2 * 23.967]
= [3.846 , 84.614]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [3.846 , 84.614] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion
c.
in this context,
type 1 error is possible because it reject the null hypothesis,
type 2 error is possible only when its fails to reject the null hypothesis.