STA2023
Statistics
EXAM Q & A
2024
, Q1. A researcher wants to compare the mean scores of two groups of students on a
standardized test. The researcher randomly assigns 50 students to group A and 50 students to
group B, and administers the test to both groups. The researcher then calculates the mean and
standard deviation of each group, and performs a t-test to test the null hypothesis that the mean
scores of the two groups are equal. The results are as follows:
Group A: mean = 75, standard deviation = 10
Group B: mean = 80, standard deviation = 15
t(98) = -2.01, p = 0.048
Which of the following statements is true based on these results?
a) The researcher can reject the null hypothesis and conclude that there is a significant
difference between the mean scores of the two groups.
b) The researcher can fail to reject the null hypothesis and conclude that there is no significant
difference between the mean scores of the two groups.
c) The researcher can reject the null hypothesis and conclude that group B has a significantly
higher mean score than group A.
d) The researcher can reject the null hypothesis and conclude that group A has a significantly
higher mean score than group B.
*Answer: c)*
Rationale: The p-value of 0.048 is less than the common significance level of 0.05, which means
that the probability of obtaining a t-value of -2.01 or more extreme under the null hypothesis is
very low. Therefore, the researcher can reject the null hypothesis and infer that there is a
statistically significant difference between the mean scores of the two groups. Since the t-value
is negative, it indicates that group A has a lower mean score than group B, and since the
absolute value of the t-value is greater than 1.96 (the critical value for a two-tailed t-test with 98
degrees of freedom at 0.05 significance level), it indicates that the difference is large enough to
be meaningful. Therefore, option c) is correct, while options a), b) and d) are incorrect.
Q2. A marketing analyst wants to estimate the proportion of customers who are satisfied with a
new product. The analyst surveys a random sample of 100 customers who bought the product,
and finds that 72 of them are satisfied. The analyst then constructs a 95% confidence interval for
the true proportion of satisfied customers in the population.
Which of the following formulas can be used to calculate the margin of error for this confidence
interval?
a) margin of error = z* * sqrt(p * (1-p) / n)
b) margin of error = t* * sqrt(p * (1-p) / n)
c) margin of error = z* * sqrt(p * q / n)
d) margin of error = t* * sqrt(p * q / n)
*Answer: a)*
Rationale: The margin of error for a confidence interval for a proportion is given by z* * sqrt(p *
Statistics
EXAM Q & A
2024
, Q1. A researcher wants to compare the mean scores of two groups of students on a
standardized test. The researcher randomly assigns 50 students to group A and 50 students to
group B, and administers the test to both groups. The researcher then calculates the mean and
standard deviation of each group, and performs a t-test to test the null hypothesis that the mean
scores of the two groups are equal. The results are as follows:
Group A: mean = 75, standard deviation = 10
Group B: mean = 80, standard deviation = 15
t(98) = -2.01, p = 0.048
Which of the following statements is true based on these results?
a) The researcher can reject the null hypothesis and conclude that there is a significant
difference between the mean scores of the two groups.
b) The researcher can fail to reject the null hypothesis and conclude that there is no significant
difference between the mean scores of the two groups.
c) The researcher can reject the null hypothesis and conclude that group B has a significantly
higher mean score than group A.
d) The researcher can reject the null hypothesis and conclude that group A has a significantly
higher mean score than group B.
*Answer: c)*
Rationale: The p-value of 0.048 is less than the common significance level of 0.05, which means
that the probability of obtaining a t-value of -2.01 or more extreme under the null hypothesis is
very low. Therefore, the researcher can reject the null hypothesis and infer that there is a
statistically significant difference between the mean scores of the two groups. Since the t-value
is negative, it indicates that group A has a lower mean score than group B, and since the
absolute value of the t-value is greater than 1.96 (the critical value for a two-tailed t-test with 98
degrees of freedom at 0.05 significance level), it indicates that the difference is large enough to
be meaningful. Therefore, option c) is correct, while options a), b) and d) are incorrect.
Q2. A marketing analyst wants to estimate the proportion of customers who are satisfied with a
new product. The analyst surveys a random sample of 100 customers who bought the product,
and finds that 72 of them are satisfied. The analyst then constructs a 95% confidence interval for
the true proportion of satisfied customers in the population.
Which of the following formulas can be used to calculate the margin of error for this confidence
interval?
a) margin of error = z* * sqrt(p * (1-p) / n)
b) margin of error = t* * sqrt(p * (1-p) / n)
c) margin of error = z* * sqrt(p * q / n)
d) margin of error = t* * sqrt(p * q / n)
*Answer: a)*
Rationale: The margin of error for a confidence interval for a proportion is given by z* * sqrt(p *
