ISDS 361A PRACTICE EXAM 2026 QUESTIONS AND
SOLUTIONS RATED A+
✔✔At α=0.10 and using the p-value
Select one:
a. We do not reject H(0)
b. We reject H(0) in favor of H(a) - ✔✔As P value < alpha. We reject H0.
b. We reject H(0) in favor of H(a)
✔✔The following 5 questions are based on this information.
In a poll of 500 Graduat students, .75% (p¯=0.75) said that they used only Internet for
the project and assignment purposes..
The goal is to construct a 99% confidence interval for the percentage (p) of Graduatel
students who use the Internet for project and assignment purposes. - ✔✔You need to
know this for the next five questions
✔✔The standard error (SE) of p¯ is
Select one:
a. 0.0004
b. 0.016
c. 0.0002
d. 0.019 - ✔✔p bar 0.75
n 500
SE 0.019365
alpha 0.01
SE= =SQRT((p bar*(1- pbar))/n
d. 0.019
✔✔The critical value (CV) needed for 99% confidence interval estimation is
Select one:
a. 2.58
b. 1.28
c. 1.64
d. 1.96 - ✔✔P value =-NORM.S.INV(0.01/2)
a. 2.58
✔✔The 99% confidence interval estimate of p is
Select one:
, a. 0.44 ± 0.002
b. 0.75 ± 0.05
c. 0.75 ± 0.15
d. 0.44 ± 0.03 - ✔✔p bar 0.75
n 500
alpha 0.01
SE 0.019365
CV 2.575829
ME 0.049881
ME = CV*SE
IE= pbar +/- ME = 0.75 +/- 0.05
b. 0.75 ± 0.05
✔✔Suppose around the period the above poll was conducted,The DEan of a university
made a personal statement saying that .85% of Graduate students used only the
Internet for assignment purposes
In light of the sample evidence and at the 1% level of significance,
Select one:
a. We cannot reject the Dean's claim
b. We can reject the Dean's claim - ✔✔Because 0.85 is not in the range 0.75 +/- 0.05.
We can reject the claim.
b. We can reject the Dean's claim
✔✔A Dean of the universityl wishes to collect new random sample with the aim of
building a new confidence interval at the 99% confidence level for p.
Using the current sample proportion (from the 500 graduate students poll ) as a basis,
what sample size (n) would the journalist require to achieve a 10% margin of error?
Select one:
a. 73
b. 500
c. 125
d. 250 - ✔✔n= z_(α/2) ^2 * p * (1-p)/E^2
c. 125
✔✔The following 5 questions are based on this information. A survey indicates that the
proportion of girls in the total youth in the United States is 51% (p=0.51).
SOLUTIONS RATED A+
✔✔At α=0.10 and using the p-value
Select one:
a. We do not reject H(0)
b. We reject H(0) in favor of H(a) - ✔✔As P value < alpha. We reject H0.
b. We reject H(0) in favor of H(a)
✔✔The following 5 questions are based on this information.
In a poll of 500 Graduat students, .75% (p¯=0.75) said that they used only Internet for
the project and assignment purposes..
The goal is to construct a 99% confidence interval for the percentage (p) of Graduatel
students who use the Internet for project and assignment purposes. - ✔✔You need to
know this for the next five questions
✔✔The standard error (SE) of p¯ is
Select one:
a. 0.0004
b. 0.016
c. 0.0002
d. 0.019 - ✔✔p bar 0.75
n 500
SE 0.019365
alpha 0.01
SE= =SQRT((p bar*(1- pbar))/n
d. 0.019
✔✔The critical value (CV) needed for 99% confidence interval estimation is
Select one:
a. 2.58
b. 1.28
c. 1.64
d. 1.96 - ✔✔P value =-NORM.S.INV(0.01/2)
a. 2.58
✔✔The 99% confidence interval estimate of p is
Select one:
, a. 0.44 ± 0.002
b. 0.75 ± 0.05
c. 0.75 ± 0.15
d. 0.44 ± 0.03 - ✔✔p bar 0.75
n 500
alpha 0.01
SE 0.019365
CV 2.575829
ME 0.049881
ME = CV*SE
IE= pbar +/- ME = 0.75 +/- 0.05
b. 0.75 ± 0.05
✔✔Suppose around the period the above poll was conducted,The DEan of a university
made a personal statement saying that .85% of Graduate students used only the
Internet for assignment purposes
In light of the sample evidence and at the 1% level of significance,
Select one:
a. We cannot reject the Dean's claim
b. We can reject the Dean's claim - ✔✔Because 0.85 is not in the range 0.75 +/- 0.05.
We can reject the claim.
b. We can reject the Dean's claim
✔✔A Dean of the universityl wishes to collect new random sample with the aim of
building a new confidence interval at the 99% confidence level for p.
Using the current sample proportion (from the 500 graduate students poll ) as a basis,
what sample size (n) would the journalist require to achieve a 10% margin of error?
Select one:
a. 73
b. 500
c. 125
d. 250 - ✔✔n= z_(α/2) ^2 * p * (1-p)/E^2
c. 125
✔✔The following 5 questions are based on this information. A survey indicates that the
proportion of girls in the total youth in the United States is 51% (p=0.51).
