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Question 1
A CDC report found arthritis among adults: 51 of 100 men and 80 of 782 women. To estimate the difference in proportions with 95% confidence, which method would you use?
Correct Answer:
2-proportion z-interval
Explanation:
Estimating the difference between two independent proportions with a confidence interval. When you want a 95% confidence interval for p1 − p2 based on two independent samples, you use a two-proportion z-interval. Take the sample proportions and compute the standard error as SE = sqrt[p1̄ (1−p1̄ )/n1 + p2̄ (1−p2̄ )/n2]. Then use the 1.96 multiplier (for 95% confidence) to form the interval: (p1̄ − p2̄ ) ± 1.96 × SE. Here, p1̄ = 51/100 = 0.51, p2̄ = 80/782 ≈ 0.1023, and the difference is about 0.4077. The standard error is SE ≈ sqrt(0.51×0.49/100 + 0.1023×0.8977/782) ≈ sqrt(0.002499 + 0.000118) ≈ 0.0512. So the 95% confidence interval for the difference is approximately 0.4077 ± 0.1004, i.e., (0.307, 0.508). This means we’re estimating that arthritis is higher among men by about 30.7 to 50.8 percentage points in this sample. The 2-proportion z-interval is the standard method for this kind of estimate; the 2-proportion z-test would be used for testing whether the two proportions are equal, not for a confidence interval, and the other options are for different goals.
Question 2
A student weighs 10 candy bars; test whether the average weight differs from the expected value using a t-test. Which test is appropriate?
Correct Answer:
1-sample t-test with 9 degrees of freedom
Explanation:
When you want to know if a sample mean differs from a known value, you use a one-sample t-test. Here you have a single sample of ten candy bars and you’re comparing the average weight to the expected mean, with the population standard deviation unknown. With n = 10, the degrees of freedom are n − 1, which is 9, so the test uses nine degrees of freedom. The test statistic is t = (x̄ − μ0) / (s/√n), reflecting how far the observed mean is from the expected mean in units of the standard error. If the population standard deviation were known, a Z-test would be used, but that’s rare in practice. It’s not a two-sample test since there aren’t two independent groups to compare, and it’s not a paired test because there’s no pairing of measurements. So the appropriate method is a one-sample t-test with nine degrees of freedom.
Question 3
A wildlife biologist wants to determine the mean weight of adult red squirrels. She weighs 10 squirrels and records a mean of 12.32 grams and a standard deviation of 1.88 grams. Which method would be used to construct a confidence interval for the mean?
Correct Answer:
1-sample t-interval with 9 degrees of freedom
Explanation:
Estimating a population mean from a small sample when the population standard deviation is unknown uses a one-sample t-interval. Here you have ten squirrels, so the degrees of freedom are n minus one, which is nine. The confidence interval for the mean is built as the sample mean plus or minus a t critical value with nine degrees of freedom times the sample standard deviation divided by the square root of the sample size. This approach accounts for the extra uncertainty introduced by estimating sigma from the sample. The z-interval would be used only if the population standard deviation were known (or the sample were very large), which isn’t the case here. The two-sample t-interval is for estimating the difference between two independent means, and the matched-pairs t-test is for paired data, not for estimating a single mean. So a one-sample t-interval with nine degrees of freedom is the appropriate method.
Question 4
A quality-control team wants to test whether the average wait time is 6 minutes using a large sample where the population standard deviation is known. Which test is appropriate?
Correct Answer:
One-Sample Z-Test
Explanation:
Testing a single population mean when you know the population standard deviation uses a Z-test for the mean. With a known sigma, the standard normal distribution determines the critical values, and the test statistic is Z = (Xbar − mu0) / (sigma / sqrt(n)). The large sample size supports the normal approximation of the sampling distribution of the mean, making the Z approach appropriate. The one-sample t-test would be used only if sigma were unknown. The two-sample t-test compares two independent means, not a mean against a fixed value. The matched-pairs t-test is for paired measurements, not a single mean against 6.
Question 5
In a chi-square test of independence with five rows and two columns, what is the degrees of freedom?
Correct Answer:
(5-1)(2-1) = 4
Explanation:
The concept being tested is how to determine degrees of freedom for a chi-square test of independence in a contingency table. For a table with r rows and c columns, the degrees of freedom are calculated as (r−1) × (c−1). This reflects how many cell counts can vary independently once the row and column totals are fixed. With five rows and two columns, the degrees of freedom are (5−1) × (2−1) = 4 × 1 = 4. Intuitively, you can freely set the counts in four cells, and the remaining counts are determined by the row and column totals, leaving four independent pieces of information. Adding the row and column reductions would overcount the constraints, and using the total number of cells (which is 10) ignores the margins entirely, so it isn’t appropriate for calculating degrees of freedom.
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Prepare with the Identify the Inference Methods Practice Test practice quiz. This question bank includes 10 questions covering weight, determine, mean, adults, and women. Use it to review important concepts, identify knowledge gaps, and build confidence for the related exam, course, or assessment.

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Identify the Inference Methods Practice Test

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