Question

Psychologists conducted a survey of 300 high school students and their attitudes towards vaping. One of the variables of interest was the response to the question, “Are you confident that you can resist peer pressure from your friends to vape?” Each response was measured on a 7-point likert scale, from 1=“Absolutely not” to 7=“Absolutely yes.”, as part of a survey given to the students after attending a presentation warning of the hazards of vaping. The psychologists reported a sample mean response of 4.75 and a sample standard deviation of 1.69 for this test item. Suppose that it is known that the true mean response to this question for students who do not attend the anti-vaping presentation is μ=4.3 for this question.

Conduct a test of hypothesis to determine whether the sample of students who attended the presentation are less likely to succumb to peer pressure. Use α =.05

b.) What is the p-value associated with your findings?

c.) If you were to instead conduct this test using α=.01, does your conclusion change? Explain why or why not

Answer 1

a.
Given that,
population mean(u)=4.3
sample mean, x =4.75
standard deviation, s =1.69
number (n)=300
null, Ho: μ=4.3
alternate, H1: μ>4.3
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.65
since our test is right-tailed
reject Ho, if to > 1.65
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =4.75-4.3/(1.69/sqrt(300))
to =4.612
| to | =4.612
critical value
the value of |t α| with n-1 = 299 d.f is 1.65
we got |to| =4.612 & | t α | =1.65
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :right tail - Ha : ( p > 4.612 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: μ=4.3
alternate, H1: μ>4.3
test statistic: 4.612
critical value: 1.65
decision: reject Ho
b.
p-value: 0
we have enough evidence to support the claim that whether the sample of students who attended the presentation are less likely to succumb to peer pressure.
c.
level of significance =0.01
Given that,
population mean(u)=4.3
sample mean, x =4.75
standard deviation, s =1.69
number (n)=300
null, Ho: μ=4.3
alternate, H1: μ>4.3
level of significance, α = 0.01
from standard normal table,right tailed t α/2 =2.339
since our test is right-tailed
reject Ho, if to > 2.339
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =4.75-4.3/(1.69/sqrt(300))
to =4.612
| to | =4.612
critical value
the value of |t α| with n-1 = 299 d.f is 2.339
we got |to| =4.612 & | t α | =2.339
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :right tail - Ha : ( p > 4.612 ) = 0
hence value of p0.01 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: μ=4.3
alternate, H1: μ>4.3
test statistic: 4.612
critical value: 2.339
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that whether the sample of students who attended the presentation are less likely to succumb to peer pressure.
conclusion will not change if level of significance changes from 0.05 to 0.01