In: Statistics and Probability
1
A well-known brokerage firm executive claimed that 40% of
investors are currently confident of meeting their investment
goals. An XYZ Investor Optimism Survey, conducted over a two week
period, found that in a sample of 100 people, 37% of them said they
are confident of meeting their goals.
Test the claim that the proportion of people who are confident is
smaller than 40% at the 0.10 significance level.
The null and alternative hypothesis would be:
H0:μ=0.4
H1:μ≠0.4
H0:p≥0.4
H1:p<0.4
H0:p≤0.4
H1:p>0.4
H0:μ≥0.4
H1:μ<0.4
H0:p=0.4
H1:p≠0.4
H0:μ≤0.4
H1:μ>0.4
The test is:
two-tailed
left-tailed
right-tailed
The test statistic is: (to 3 decimals)
The p-value is: (to 4 decimals)
Based on this we:
2
20% of all college students volunteer their time. Is the percentage of college students who are volunteers smaller for students receiving financial aid? Of the 356 randomly selected students who receive financial aid, 50 of them volunteered their time. What can be concluded at the α
= 0.05 level of significance?
H0:
(please enter a decimal)
H1:
(Please enter a decimal)
a The data suggest the populaton proportion is significantly lower than 20% at α = 0.05, so there is sufficient evidence to conclude that the percentage of financial aid recipients who volunteer is lower than 20%.
b The data suggest the population proportion is not significantly lower than 20% at α = 0.05, so there is sufficient evidence to conclude that the percentage of financial aid recipients who volunteer is equal to 20%.
c The data suggest the population proportion is not significantly lower than 20% at α = 0.05, so there is insufficient evidence to conclude that the percentage of financial aid recipients who volunteer is lower than 20%.
(1)
Here we have given that,
n=Sample Size=100
=sample proportion of people confident of meeting their goals = 37%=0.37
Claim: To check whether the population proportion of people who are confident is smaller than 40% i.e. 0.40.
The null and alternative hypothesis are as follows,
Versus
where p is the population proportion of people who are confident
This is the left one-tailed test.
Now, we can find the test statistic is as follows,
Z-statistics=
=
= -0.61
The test statistics is -0.61.
Now we find the P-value,
p-value=P(Z > z-statistics)) as this is left one-tailed test
=P( Z < -0.61)
= 0.2709 Using standard normal z table see the value corresponding to the z=-0.61
The p-value is 0.2709
Decision:
= level of significance= 0.10
Here p-value (0.2709) greater than (>) 0.10
Conclusion:
We fail to reject the Ho (Null Hypothesis)
There is no sufficient evidence to support the claim the population proportion of people who are confident is smaller than 40% i.e. 0.40.
(2)
Here we have given that,
n=Sample Size=356
x= number of selected students who receive financial aid volunteered their time=50
=sample proportion of people confident in meeting their goals =
Claim: To check whether the population proportion of financial aid recipients who volunteer is lower than 20% i.e. 0.20.
The null and alternative hypothesis are as follows,
Versus
where p is the population proportion of financial aid recipients who volunteer
This is the left one-tailed test.
Now, we can find the test statistic is as follows,
Z-statistics=
=
= -2.83
The test statistics is -2.83.
Now we find the P-value,
p-value=P(Z > z-statistics)) as this is left one-tailed test
=P( Z < -2.83)
= 0.0023 Using standard normal z table see the value corresponding to the z=-0.61
The p-value is 0.0023
Decision:
= level of significance= 0.05
Here p-value (0.0023) less than (<) 0.05
Conclusion:
We reject the Ho (Null Hypothesis)
The data suggest that population proportion is significantly lower than 20% at =0.05 so there is sufficient evidence to conclude that the percentage of financial aid recipients who volunteer is lower than 20%.
i.e. option a is correct.