Question

Two professors at a local college developed a new teaching curriculum designed to increase​ students' grades in math classes. In a typical developmental math​ course,

4949​%

of the students complete the course with a letter grade of​ A, B, or C. In the experimental​ course, of the

1919

students​ enrolled,

1212

completed the course with a letter grade of​ A, B, or C. Is the experimental course effective at the

alpha equals 0.05α=0.05

level of​ significance? Complete parts ​(a) through​ (g).

​(a) State the appropriate null and alternative hypotheses.

Upper H 0H0​:

pp

pp

sigmaσ

muμ

equals=

greater than>

not equals≠

less than<

equals=

. 49.49 versus

Upper H 1H1​:

pp

sigmaσ

muμ

pp

greater than>

less than<

not equals≠

greater than>

equals=

. 49.49

​(Type integers or decimals. Do not​ round.)

​(b) Verify that the normal model may not be used to estimate the​ P-value.

Because np 0 left parenthesis 1 minus p 0 right parenthesisnp01−p0equals=4.74.7

less than<

not equals≠

equals=

greater than>

less than<

​10, the normal model

may not

may not

may

be used to approximate the​ P-value.

​(Round to one decimal place as​ needed.)

​(c) Explain why this is a binomial experiment.

There is a

fixed

variable

fixed

number of trials with

two mutually exclusive outcomes.

two mutually exclusive outcomes.

one outcome.

three outcomes.

two outcomes that can occur simultaneously.

The trials

are

are not

are

independent and the probability of success is fixed at

. 49.49

for each trial.

​(Type an integer or a decimal. Do not​ round.)

​(d) Determine the​ P-value using the binomial probability distribution. State your conclusion to the hypothesis test.

First determine the​ P-value.

​P-valueequals=. 158.158

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

Is there sufficient evidence to support the research that the experimental course is​ effective?

A.

YesYes​,

do not rejectdo not reject

the null hypothesis because the​ P-value is

greatergreater

than

alphaα.

There

is sufficientis sufficient

evidence to conclude that the experimental course is effective.

B.

NoNo​,

do not rejectdo not reject

the null hypothesis because the​ P-value is

greatergreater

than

alphaα.

There

is insufficientis insufficient

evidence to conclude that the experimental course is effective.

C.

NoNo​,

rejectreject

the null hypothesis because the​ P-value is

lessless

than

alphaα.

There

is insufficientis insufficient

evidence to conclude that the experimental course is effective.

D.

YesYes​,

rejectreject

the null hypothesis because the​ P-value is

lessless

than

alphaα.

There

is sufficientis sufficient

evidence to conclude that the experimental course is effective.​(e) Suppose the course is taught with

5757

students and

3636

complete the course with a letter grade of​ A, B, or C. Verify whether the normal model may now be used to estimate the​ P-value.Because np 0 left parenthesis 1 minus p 0 right parenthesisnp01−p0equals=14.214.2

greater than>

greater than>

equals=

less than<

not equals≠

​10, the sample size is

less than

greater than

less than

​5% of the population​ size, and the sample

can be reasonably assumed to be random,

is given to be random,

cannot be reasonably assumed to be random,

is given to not be random,

can be reasonably assumed to be random,

the normal model

may

may not

may

be used to approximate the​ P-value.

​(Round to one decimal place as​ needed.)

​(f) Use the normal model to obtain and interpret the​ P-value. State your conclusion to the hypothesis test.

First find the test​ statistic,

z 0z0.

z 0z0equals=2.62.6

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

Now determine the​ P-value.

​P-valueequals=. 005.005

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

Is there sufficient evidence to support the research that the experimental course is​ effective?

A.

NoNo​,

do not rejectdo not reject

the null hypothesis because the​ P-value is

greatergreater

than

alphaα.

There is

insufficientinsufficient

evidence to conclude that the experimental course is effective.

B.

NoNo​,

do not rejectdo not reject

the null hypothesis because the​ P-value is

lessless

than

alphaα.

There is

insufficientinsufficient

evidence to conclude that the experimental course is effective.

C.

YesYes​,

rejectreject

the null hypothesis because the​ P-value is

lessless

than

alphaα.

There is

sufficientsufficient

evidence to conclude that the experimental course is effective.

D.

YesYes​,

rejectreject

the null hypothesis because the​ P-value is

greatergreater

than

alphaα.

There is

sufficientsufficient

evidence to conclude that the experimental course is effective.

​(g) Explain the role that sample size plays in the ability to reject statements in the null hypothesis.

When there are small sample​ sizes, the evidence against the statement in the null hypothesis must be

substantial.

insignificant.

substantial.

One should be wary of studies that

reject

do not reject

reject

the null hypothesis when the test was conducted with a small sample size.

Answer 1

(a)
H0: p = 0.49
versus
H1: p > 0.49

(b)

np = 19 * 0.49 = 9.31
n(1-p) = 19 * (1 - 0.49) = 9.69
Because np and n(1-p) < 10, normal model may not be used to approximate the​ P-value.

(c)
There is a fixed number of trials with two mutually exclusive outcomes.
The trials are independent and the probability of success is fixed at 0.49 for each trial.

(d)
P-value = P(X > 12, n = 19, p = 0.49) = 0.158 (By Binomial distribution calculator)
Since, p-value is greater than 0.05 significance level, we fail to reject null hypothesis H0 and conclude that there is no strong evidence that p > 0.49.

B.

No​, do not reject the null hypothesis because the​ P-value is greater than alpha α.
There is insufficient evidence to conclude that the experimental course is effective.

(e)
np = 57 * 0.49 = 27.93
n(1-p) = 57 * (1 - 0.49) = 29.07
Because np and n(1-p) > 10,
the sample size is less than​5% of the population​ size, and the sample can be reasonably assumed to be random,
the normal model may be used to approximate the​ P-value.

(f)
Mean = 57 * 0.49 = 27.93
Standard deviation = sqrt(n * p * (1-p)) = sqrt(57 * 0.49 * (1-0.49)) = 3.774
z0 = (36 - 27.93) / 3.774 = 2.14

P-value = P(z > 2.14) = 0.016

C.

Yes,reject the null hypothesis because the​ P-value is less than alpha α.
There is sufficient evidence to conclude that the experimental course is effective.

(g)
When there are small sample​ sizes, the evidence against the statement in the null hypothesis must be substantial.

One should be wary of studies that do not reject the null hypothesis when the test was conducted with a small sample size.