ISDS 361A FINAL 2026 EXAM PAPER QUESTIONS AND
SOLUTIONS SCORED A+
✔✔In hypothesis testing, the level of significance, α, represents - ✔✔The probability of
making Type I error
✔✔An error that occurs when we fail to reject H(0) while false is called - ✔✔Type II error
✔✔Which one of the following cannot be an alternative hypothesis H(a)? - ✔✔p ≥ 0.3
✔✔Questions 6 to 10 are related. Based on a random sample of n=15 observations, we
have obtained a sample mean of x(bar)=1050. The goal is to test:
H(0): μ≤1000 and
H(a): μ>1000
Assume that x is normally distributed with σ=150. What is the standard error (σ (x bar)
)? Rounded to two decimal places. - ✔✔150/√15 = 38.73
✔✔To conduct the test as given in question 6), which test statistics, is appropriate? -
✔✔z
✔✔Building on question 7), what is the value of the test statistics? Rounded to two
decimal places. - ✔✔(1050-1000)/38.73 = 1.29
✔✔Using α=0.1, what is the critical value (use Excel) for testing the hypotheses in
question 6)? Rounded to two decimal places. - ✔✔-NORM.S.INV(0.1) = 1.28
✔✔Using correct answers for questions 8) and 9), - ✔✔We reject H(0) because the test
statistics value is greater than the critical value
✔✔Question 11-14 are related. Based on the random sample of n=800 observations,
we have obtained a sample proportion p(bar) =0.44. The goal is to test:
H(0): p≥0.48
H9a): p<0.48
What is the standard error (σ(p))? Rounded to four decimal places. -
✔✔√((0.48*0.52)/800) = 0.0177
✔✔What is the value of the test statistics? Rounded to two decimal places. -
✔✔z=(0.44-0.48)/0.0177 = -2.26
✔✔Using α=0.05, what is the critical value (use Excel) for testing the hypotheses in
question 11)? Rounded to two decimal places. - ✔✔NORM.S.INV (0.05) = -1.64
,✔✔Using correct answers for questions 12) and 13), - ✔✔We reject H(0)because z is
less than the critical value
✔✔The following 5 questions are based on this information: An economist claims that
average weekly food expenditure of households in City 1 is more than that of
households in City 2. She surveys 35 households in City 1 and obtains an average
weekly food expenditure of $164. A sample of 30 households in City 2 yields an
average weekly expenditure of $159. Historical data reveals that the populationstandard
deviation for City 1 and City 2 are $12.50 and $9.25, respectively.
City 1
City 2
x1(bar)=164
x2 (bar) =159
σ(1)=12.5
σ(2) =9.25
n(1)=35
n2=30
Let μ(1) be the mean weekly food expenditure for City 1 and μ(2) be that for City 2. -
✔✔
✔✔The standard error of x(1)bar- x(2) bar is - ✔✔2.70
✔✔The value of the test statistics is - ✔✔1.85
✔✔The p-value of the test is - ✔✔0.03
✔✔At α=0.05 - ✔✔We can reject H(0) in favor of H(a)
✔✔The following 8 questions are based on this information.
The table below shows the annual return data for 10 firms in the gold industry and 10
firms in the oil industry. Here we assume that these two samples are from tow normal
populations. Annual returns (in percent)
Gold
Oil
6
-3
15
15
19
28
, 26
18
2
32
16
31
31
15
14
12
15
10
16
15
Let μ(1) be the mean return for the gold industry and μ(2) be the mean return for the oil
industry. We wish to test if the average returns in the two industries are different? Here
we assume that these two samples are from two independent normal populations. - ✔✔
✔✔The competing hypotheses should be formulated as - ✔✔H(0): μ(1)-μ(2)=0 versus
H(a):μ(1)-μ(2)≠0
✔✔The test statistics value is - ✔✔-0.3023
✔✔The p-value of the test is - ✔✔0.7661
✔✔At α=0.1, - ✔✔We cannot reject H(0)
✔✔What is the standard error of (x(1)bar- x(2)bar? - ✔✔4.30
✔✔Suppose you would like to construct the 90% confidence interval for μ(1)-μ(2). The
critical value, t(α/2) is - ✔✔1.74
✔✔The 90% confidence interval for μ(1)-μ(2) is - ✔✔-1.3 ±7.48
✔✔Using the 90% confidence interval estimate, we do not reject H_0 because - ✔✔0 is
included in the interval
✔✔The standard deviation of the sampling distribution of the sample mean is also
called - ✔✔standard error
✔✔The mean of the sampling distribution of the sample mean is: - ✔✔equal to the
population mean
SOLUTIONS SCORED A+
✔✔In hypothesis testing, the level of significance, α, represents - ✔✔The probability of
making Type I error
✔✔An error that occurs when we fail to reject H(0) while false is called - ✔✔Type II error
✔✔Which one of the following cannot be an alternative hypothesis H(a)? - ✔✔p ≥ 0.3
✔✔Questions 6 to 10 are related. Based on a random sample of n=15 observations, we
have obtained a sample mean of x(bar)=1050. The goal is to test:
H(0): μ≤1000 and
H(a): μ>1000
Assume that x is normally distributed with σ=150. What is the standard error (σ (x bar)
)? Rounded to two decimal places. - ✔✔150/√15 = 38.73
✔✔To conduct the test as given in question 6), which test statistics, is appropriate? -
✔✔z
✔✔Building on question 7), what is the value of the test statistics? Rounded to two
decimal places. - ✔✔(1050-1000)/38.73 = 1.29
✔✔Using α=0.1, what is the critical value (use Excel) for testing the hypotheses in
question 6)? Rounded to two decimal places. - ✔✔-NORM.S.INV(0.1) = 1.28
✔✔Using correct answers for questions 8) and 9), - ✔✔We reject H(0) because the test
statistics value is greater than the critical value
✔✔Question 11-14 are related. Based on the random sample of n=800 observations,
we have obtained a sample proportion p(bar) =0.44. The goal is to test:
H(0): p≥0.48
H9a): p<0.48
What is the standard error (σ(p))? Rounded to four decimal places. -
✔✔√((0.48*0.52)/800) = 0.0177
✔✔What is the value of the test statistics? Rounded to two decimal places. -
✔✔z=(0.44-0.48)/0.0177 = -2.26
✔✔Using α=0.05, what is the critical value (use Excel) for testing the hypotheses in
question 11)? Rounded to two decimal places. - ✔✔NORM.S.INV (0.05) = -1.64
,✔✔Using correct answers for questions 12) and 13), - ✔✔We reject H(0)because z is
less than the critical value
✔✔The following 5 questions are based on this information: An economist claims that
average weekly food expenditure of households in City 1 is more than that of
households in City 2. She surveys 35 households in City 1 and obtains an average
weekly food expenditure of $164. A sample of 30 households in City 2 yields an
average weekly expenditure of $159. Historical data reveals that the populationstandard
deviation for City 1 and City 2 are $12.50 and $9.25, respectively.
City 1
City 2
x1(bar)=164
x2 (bar) =159
σ(1)=12.5
σ(2) =9.25
n(1)=35
n2=30
Let μ(1) be the mean weekly food expenditure for City 1 and μ(2) be that for City 2. -
✔✔
✔✔The standard error of x(1)bar- x(2) bar is - ✔✔2.70
✔✔The value of the test statistics is - ✔✔1.85
✔✔The p-value of the test is - ✔✔0.03
✔✔At α=0.05 - ✔✔We can reject H(0) in favor of H(a)
✔✔The following 8 questions are based on this information.
The table below shows the annual return data for 10 firms in the gold industry and 10
firms in the oil industry. Here we assume that these two samples are from tow normal
populations. Annual returns (in percent)
Gold
Oil
6
-3
15
15
19
28
, 26
18
2
32
16
31
31
15
14
12
15
10
16
15
Let μ(1) be the mean return for the gold industry and μ(2) be the mean return for the oil
industry. We wish to test if the average returns in the two industries are different? Here
we assume that these two samples are from two independent normal populations. - ✔✔
✔✔The competing hypotheses should be formulated as - ✔✔H(0): μ(1)-μ(2)=0 versus
H(a):μ(1)-μ(2)≠0
✔✔The test statistics value is - ✔✔-0.3023
✔✔The p-value of the test is - ✔✔0.7661
✔✔At α=0.1, - ✔✔We cannot reject H(0)
✔✔What is the standard error of (x(1)bar- x(2)bar? - ✔✔4.30
✔✔Suppose you would like to construct the 90% confidence interval for μ(1)-μ(2). The
critical value, t(α/2) is - ✔✔1.74
✔✔The 90% confidence interval for μ(1)-μ(2) is - ✔✔-1.3 ±7.48
✔✔Using the 90% confidence interval estimate, we do not reject H_0 because - ✔✔0 is
included in the interval
✔✔The standard deviation of the sampling distribution of the sample mean is also
called - ✔✔standard error
✔✔The mean of the sampling distribution of the sample mean is: - ✔✔equal to the
population mean
