Question

In a random sample of​ males, it was found that

26

write with their left hands and

214

do not. In a random sample of​ females, it was found that

62

write with their left hands and

438

do not. Use a

0.05

significance level to test the claim that the rate of​ left-handedness among males is less than that among females. Complete parts​ (a) through​ (c) below.

a. Test the claim using a hypothesis test.

Consider the first sample to be the sample of males and the second sample to be the sample of females. What are the null and alternative hypotheses for the hypothesis​ test?

A.

Upper H 0

​:

p 1

greater than or equalsp 2

Upper H 1

​:

p 1

not equalsp 2

B.

Upper H 0

​:

p 1

not equalsp 2

Upper H 1

​:

p 1

equalsp 2

C.

Upper H 0

​:

p 1

equalsp 2

Upper H 1

​:

p 1

not equalsp 2

D.

Upper H 0

​:

p 1

equalsp 2

Upper H 1

​:

p 1

greater thanp 2

E.

Upper H 0

​:

p 1

less than or equalsp 2

Upper H 1

​:

p 1

not equalsp 2

F.

Upper H 0

​:

p 1

equalsp 2

Upper H 1

​:

p 1

less thanp 2

Identify the test statistic.

zequals

nothing

​(Round to two decimal places as​ needed.)

Identify the​ P-value.

​P-valueequals

nothing

​(Round to three decimal places as​ needed.)

What is the conclusion based on the hypothesis​ test?

The​ P-value is

▼

less than

greater than

the significance level of

alpha

equals0.05

​,

so

▼

reject

fail to reject

the null hypothesis. There

▼

is sufficient

is not sufficient

evidence to support the claim that the rate of​ left-handedness among males is less than that among females.

b. Test the claim by constructing an appropriate confidence interval.

The

90

​%

confidence interval is

nothing

less thanleft parenthesis p 1 minus p 2 right parenthesisless thannothing

.

​(Round to three decimal places as​ needed.)

What is the conclusion based on the confidence​ interval?

Because the confidence interval limits

▼

do not include

include

​0, it appears that the two rates of​ left-handedness are

▼

not equal.

equal.

There

▼

is sufficient

is not sufficient

evidence to support the claim that the rate of​ left-handedness among males is less than that among females.

c. Based on the​ results, is the rate of​ left-handedness among males less than the rate of​ left-handedness among​ females?

A.

The rate of​ left-handedness among males

does

appear to be less than the rate of​ left-handedness among females because the results are statistically significant.

B.

The rate of​ left-handedness among males

does

appear to be less than the rate of​ left-handedness among females because the results are not statistically significant.

C.

The rate of​ left-handedness among males

does not

appear to be less than the rate of​ left-handedness among females.

D.

The results are inconclusive.

Answer 1

p1cap = X1/N1 = 26/214 = 0.1215
p1cap = X2/N2 = 62/438 = 0.1416
pcap = (X1 + X2)/(N1 + N2) = (26+62)/(214+438) = 0.135

a)
Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p1 = p2
Alternate Hypothesis, Ha: p1 < p2

Test statistic
z = (p1cap - p2cap)/sqrt(pcap * (1-pcap) * (1/N1 + 1/N2))
z = (0.1215-0.1416)/sqrt(0.135*(1-0.135)*(1/214 + 1/438))
z = -0.71

P-value Approach
P-value = 0.239

As P-value >= 0.05, fail to reject null hypothesis.
There is not sufficient evidence to conclude that the left handedness in males is less than females.

b)
Here, , n1 = 214 , n2 = 438
p1cap = 0.1215 , p2cap = 0.1416

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.1215 * (1-0.1215)/214 + 0.1416*(1-0.1416)/438)
SE = 0.0279

For 0.9 CI, z-value = 1.64
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.1215 - 0.1416 - 1.64*0.0279, 0.1215 - 0.1416 + 1.64*0.0279)
CI = (-0.066 , 0.026)

Because CI interval includes 0, it apprears that the two rates of left handedness are equal

There is not sufficient evidence

c)
option C