Statistical Inference Quiz Answers – Complete Graded Quiz

Welcome to your comprehensive guide for Statistical Inference quiz answers! Whether you’re tackling practice quizzes to reinforce your understanding or preparing for graded quizzes to assess your knowledge, this guide is here to help.

Covering all course modules, this resource will help you master core statistical inference concepts, including hypothesis testing, confidence intervals, p-values, and statistical decision-making.

Statistical Inference Quiz Answers for All Modules

Statistical Inference Week 01 Quiz Answers

Q1. Consider influenza epidemics for two parent heterosexual families. Suppose that the probability is 17% that at least one of the parents has contracted the disease. The probability that the father has contracted influenza is 12% while the probability that both the mother and father have contracted the disease is 6%. What is the probability that the mother has contracted influenza?

The correct answer is 11%.

Explanation: To calculate the probability that the mother has contracted influenza, use the formula for the union of two events: P(Father or Mother)=P(Father)+P(Mother)−P(Father and Mother)P(\text{Father or Mother}) = P(\text{Father}) + P(\text{Mother}) – P(\text{Father and Mother})P(Father or Mother)=P(Father)+P(Mother)−P(Father and Mother) Given:

P(Father)=0.12P(\text{Father}) = 0.12P(Father)=0.12
P(Father or Mother)=0.17P(\text{Father or Mother}) = 0.17P(Father or Mother)=0.17
P(Father and Mother)=0.06P(\text{Father and Mother}) = 0.06P(Father and Mother)=0.06

Thus, P(Mother)=0.17−0.12+0.06=0.11P(\text{Mother}) = 0.17 – 0.12 + 0.06 = 0.11P(Mother)=0.17−0.12+0.06=0.11.

Q2. A random variable, X is uniform, a box from 0 to 1 of height 1. (So that its density is f(x)=1f(x) = 1f(x)=1 for 0≤x≤10 \leq x \leq 10≤x≤1). What is its 75th percentile?

The correct answer is 0.75.

Explanation: The cumulative distribution function (CDF) for a uniform distribution from 0 to 1 is F(x)=xF(x) = xF(x)=x. To find the 75th percentile, set F(x)=0.75F(x) = 0.75F(x)=0.75, so x=0.75x = 0.75x=0.75.

Q3. You are playing a game with a friend where you flip a coin and if it comes up heads you give her X dollars and if it comes up tails she gives you Y dollars. The probability that the coin is heads is p (some number between 0 and 1.) What has to be true about X and Y to make so that both of your expected total earnings is 0. The game would then be called “fair”.

The correct answer is p(1−p)=X/Yp(1 – p) = X/Yp(1−p)=X/Y.

Explanation: The expected total earnings for each player must be zero. For the first player, the expected earnings are p×X−(1−p)×Yp \times X – (1 – p) \times Yp×X−(1−p)×Y. Setting this equal to zero, we get the condition p(1−p)=X/Yp(1 – p) = X/Yp(1−p)=X/Y.

Q4. A density that looks like a normal density (but may or may not be exactly normal) is exactly symmetric about zero. (Symmetric means if you flip it around zero it looks the same.) What is its median?

The correct answer is 0.

Explanation: For a symmetric distribution, the median is the same as the mean. Since the distribution is symmetric about zero, the median must be 0.

Q5. Consider the following PMF shown below in R

RCopyEditx <- 1:4
p <- x/sum(x)
temp <- rbind(x, p)
rownames(temp) <- c("X", "Prob")
temp

What is the mean?

The correct answer is 2.

Explanation: The mean of a probability mass function (PMF) is calculated as the sum of x×p(x)x \times p(x)x×p(x), where xxx is the value and p(x)p(x)p(x) is its corresponding probability. The PMF here is x={1,2,3,4}x = \{1, 2, 3, 4\}x={1,2,3,4} and p(x)={0.1,0.2,0.3,0.4}p(x) = \{0.1, 0.2, 0.3, 0.4\}p(x)={0.1,0.2,0.3,0.4}. The mean is:Mean=1×0.1+2×0.2+3×0.3+4×0.4=2\text{Mean} = 1 \times 0.1 + 2 \times 0.2 + 3 \times 0.3 + 4 \times 0.4 = 2Mean=1×0.1+2×0.2+3×0.3+4×0.4=2

Q6. A website for home pregnancy tests cites the following: “When the subjects using the test were women who collected and tested their own samples, the overall sensitivity was 75%. Specificity was also low, in the range 52% to 75%.” Assume the lower value for the specificity. Suppose a subject has a positive test and that 30% of women taking pregnancy tests are actually pregnant. What number is closest to the probability of pregnancy given the positive test?

The correct answer is 20%.

Explanation: To apply Bayes’ rule, we calculate:P(Pregnant∣Positive)=P(Positive∣Pregnant)×P(Pregnant)P(Positive)P(\text{Pregnant} | \text{Positive}) = \frac{P(\text{Positive} | \text{Pregnant}) \times P(\text{Pregnant})}{P(\text{Positive})}P(Pregnant∣Positive)=P(Positive)P(Positive∣Pregnant)×P(Pregnant)

Where:

P(Positive∣Pregnant)=0.75P(\text{Positive} | \text{Pregnant}) = 0.75P(Positive∣Pregnant)=0.75 (sensitivity)
P(Pregnant)=0.30P(\text{Pregnant}) = 0.30P(Pregnant)=0.30
P(Positive)=P(Positive∣Pregnant)×P(Pregnant)+P(Positive∣Not Pregnant)×P(Not Pregnant)P(\text{Positive}) = P(\text{Positive} | \text{Pregnant}) \times P(\text{Pregnant}) + P(\text{Positive} | \text{Not Pregnant}) \times P(\text{Not Pregnant})P(Positive)=P(Positive∣Pregnant)×P(Pregnant)+P(Positive∣Not Pregnant)×P(Not Pregnant)

Plugging in the values, we get P(Pregnant∣Positive)≈0.20P(\text{Pregnant} | \text{Positive}) \approx 0.20P(Pregnant∣Positive)≈0.20.

Statistical Inference Week 02 Quiz Answers

Q1. What is the variance of the distribution of the average an IID draw of nnn observations from a population with mean μ\muμ and variance σ2\sigma^2σ2?

The correct answer is σ2/n\sigma^2/nσ2/n.

Explanation: The variance of the sample mean for nnn independent and identically distributed (IID) observations is the population variance divided by nnn. That is, the variance of the sample mean is σ2n\frac{\sigma^2}{n}nσ2.

Q2. Suppose that diastolic blood pressures (DBPs) for men aged 35-44 are normally distributed with a mean of 80 (mm Hg) and a standard deviation of 10. About what is the probability that a random 35-44 year old has a DBP less than 70?

The correct answer is 16%.

Explanation: To find the probability that a randomly selected individual has a DBP less than 70, we calculate the z-score:z=X−μσ=70−8010=−1z = \frac{X – \mu}{\sigma} = \frac{70 – 80}{10} = -1z=σX−μ=1070−80=−1

Using the standard normal distribution table, a z-score of -1 corresponds to a probability of approximately 0.1587, or about 16%.

Q3. Brain volume for adult women is normally distributed with a mean of about 1,100 cc for women with a standard deviation of 75 cc. What brain volume represents the 95th percentile?

The correct answer is approximately 1223.

Explanation: The 95th percentile corresponds to a z-score of approximately 1.645. Using the formula for the z-score:X=μ+z⋅σ=1100+1.645⋅75=1223.875X = \mu + z \cdot \sigma = 1100 + 1.645 \cdot 75 = 1223.875X=μ+z⋅σ=1100+1.645⋅75=1223.875

Rounding this gives approximately 1223 cc.

Q4. Refer to the previous question. Brain volume for adult women is about 1,100 cc for women with a standard deviation of 75 cc. Consider the sample mean of 100 random adult women from this population. What is the 95th percentile of the distribution of that sample mean?

The correct answer is approximately 1112 cc.

Explanation: For the sample mean, the standard deviation of the sample mean is σn=75100=7.5\frac{\sigma}{\sqrt{n}} = \frac{75}{\sqrt{100}} = 7.5nσ=10075=7.5. For the 95th percentile, we use a z-score of approximately 1.645:X=μ+z⋅SE=1100+1.645⋅7.5=1112.875X = \mu + z \cdot \text{SE} = 1100 + 1.645 \cdot 7.5 = 1112.875X=μ+z⋅SE=1100+1.645⋅7.5=1112.875

Rounding this gives approximately 1112 cc.

Q5. You flip a fair coin 5 times, about what’s the probability of getting 4 or 5 heads?

The correct answer is 12%.

Explanation: The probability of getting exactly 4 heads is:P(X=4)=(54)(12)5=5⋅132=532P(X = 4) = \binom{5}{4} \left(\frac{1}{2}\right)^5 = 5 \cdot \frac{1}{32} = \frac{5}{32}P(X=4)=(45)(21)5=5⋅321=325

The probability of getting exactly 5 heads is:P(X=5)=(55)(12)5=1⋅132=132P(X = 5) = \binom{5}{5} \left(\frac{1}{2}\right)^5 = 1 \cdot \frac{1}{32} = \frac{1}{32}P(X=5)=(55)(21)5=1⋅321=321

Thus, the total probability of getting 4 or 5 heads is:P(X=4 or X=5)=532+132=632=0.1875 or 12%P(X = 4 \text{ or } X = 5) = \frac{5}{32} + \frac{1}{32} = \frac{6}{32} = 0.1875 \text{ or } 12\%P(X=4 or X=5)=325+321=326=0.1875 or 12%

Q6. The respiratory disturbance index (RDI), a measure of sleep disturbance, for a specific population has a mean of 15 (sleep events per hour) and a standard deviation of 10. They are not normally distributed. Give your best estimate of the probability that a sample mean RDI of 100 people is between 14 and 16 events per hour?

The correct answer is 68%.

Explanation: By the Central Limit Theorem, the sampling distribution of the sample mean is approximately normal for a large enough sample size. With a sample size of 100, the standard deviation of the sample mean is:SE=10100=1\text{SE} = \frac{10}{\sqrt{100}} = 1SE=10010=1

For a sample mean between 14 and 16, we are looking at a range of approximately 1 standard deviation, which corresponds to about 68% of the values in a normal distribution.

Q7. Consider a standard uniform density. The mean for this density is 0.5 and the variance is 1/12. You sample 1,000 observations from this distribution and take the sample mean, what value would you expect it to be near?

The correct answer is 0.5.

Explanation: The mean of the standard uniform distribution is 0.5. Since the sample mean is an unbiased estimator of the population mean, the expected value of the sample mean will be near 0.5.

Q8. The number of people showing up at a bus stop is assumed to be Poisson with a mean of 5 people per hour. You watch the bus stop for 3 hours. About what’s the probability of viewing 10 or fewer people?

The correct answer is 0.08.

Explanation: The Poisson distribution with a mean of 5 people per hour for 3 hours has a mean of 5×3=155 \times 3 = 155×3=15. Using the cumulative distribution function for the Poisson distribution, the probability of seeing 10 or fewer people is approximately 0.08.

Statistical Inference Week 03 Quiz Answers

Q1. What is the variance of the distribution of the average an IID draw of nnn observations from a population with mean μ\muμ and variance σ2\sigma^2σ2?

The correct answer is σ2n\frac{\sigma^2}{n}nσ2.

Explanation: The variance of the sample mean for nnn independent and identically distributed (IID) observations is the population variance divided by nnn. Thus, the variance of the sample mean is σ2n\frac{\sigma^2}{n}nσ2.

The correct answer is 16%.

Explanation: The z-score for DBP = 70 is:z=70−8010=−1z = \frac{70 – 80}{10} = -1z=1070−80=−1

Using the standard normal distribution table, the probability for a z-score of -1 is approximately 0.1587, or about 16%.

Q3. Brain volume for adult women is normally distributed with a mean of about 1,100 cc for women with a standard deviation of 75 cc. What brain volume represents the 95th percentile?

The correct answer is approximately 1223.

Explanation: The 95th percentile corresponds to a z-score of 1.645. Using the z-score formula:X=μ+z⋅σ=1100+1.645⋅75=1223.875X = \mu + z \cdot \sigma = 1100 + 1.645 \cdot 75 = 1223.875X=μ+z⋅σ=1100+1.645⋅75=1223.875

Rounding gives approximately 1223 cc.

The correct answer is approximately 1112 cc.

Explanation: For the sample mean, the standard error is 75100=7.5\frac{75}{\sqrt{100}} = 7.510075=7.5. Using a z-score of 1.645 for the 95th percentile:X=μ+z⋅SE=1100+1.645⋅7.5=1112.875X = \mu + z \cdot \text{SE} = 1100 + 1.645 \cdot 7.5 = 1112.875X=μ+z⋅SE=1100+1.645⋅7.5=1112.875

Rounding gives approximately 1112 cc.

Q5. You flip a fair coin 5 times, about what’s the probability of getting 4 or 5 heads?

The correct answer is 12%.

Explanation: The probability of getting 4 heads is:P(X=4)=(54)(12)5=5⋅132=532P(X = 4) = \binom{5}{4} \left(\frac{1}{2}\right)^5 = 5 \cdot \frac{1}{32} = \frac{5}{32}P(X=4)=(45)(21)5=5⋅321=325

The probability of getting 5 heads is:P(X=5)=(55)(12)5=1⋅132=132P(X = 5) = \binom{5}{5} \left(\frac{1}{2}\right)^5 = 1 \cdot \frac{1}{32} = \frac{1}{32}P(X=5)=(55)(21)5=1⋅321=321

The correct answer is 68%.

Explanation: By the Central Limit Theorem, the sampling distribution of the sample mean will be approximately normal with a standard deviation of:SE=10100=1\text{SE} = \frac{10}{\sqrt{100}} = 1SE=10010=1

For a sample mean between 14 and 16, we are looking at a range of approximately 1 standard deviation, which corresponds to 68% of the values in a normal distribution.

The correct answer is 0.5.

Explanation: The mean of the standard uniform distribution is 0.5. The sample mean is an unbiased estimator of the population mean, so we expect the sample mean to be close to 0.5.

The correct answer is 0.08.

Statistical Inference Week 04 Quiz Answers

Q1. What is the P-value for the associated two sided T test for testing the hypothesis of a mean reduction in blood pressure for the given data?

The correct answer is 0.043.

Explanation: To perform a paired t-test on the blood pressure data, we calculate the differences between the baseline and week 2 values, then compute the t-statistic. The P-value indicates the probability of observing the data assuming the null hypothesis (no difference in blood pressure). In this case, the P-value is 0.043, meaning there is evidence to suggest a mean reduction in blood pressure.

Q2. What is the complete set of values of μ0\mu_0μ0 that a test of H0:μ=μ0H_0: \mu = \mu_0H0:μ=μ0 would fail to reject the null hypothesis in a two sided 5% Student’s t-test?

The correct answer is 1077 to 1123.

Explanation: The confidence interval for the sample mean (1,100 ± 2 times the standard error) will give the range of μ0\mu_0μ0 values for which the null hypothesis is not rejected. The standard error is calculated as 309\frac{30}{\sqrt{9}}930, which gives the range 1077 to 1123 for μ0\mu_0μ0.

Q3. What is the P-value for a one-sided exact test to test if Coke is preferred to Pepsi given that 3 out of 4 people chose Coke?

The correct answer is 0.31.

Explanation: The test follows a binomial distribution with 4 trials and 3 successes. Using the binomial test, the P-value for the observed result (3 successes out of 4) is 0.31, suggesting that Coke is not significantly preferred over Pepsi at the 5% significance level.

Q4. What is the one sided P-value for the test of whether the hospital’s infection rate is below the standard of 1 infection per 100 person-days?

The correct answer is 0.03.

Explanation: The observed infection rate is 101787≈0.0056\frac{10}{1787} \approx 0.0056178710≈0.0056, which is much lower than the benchmark of 1 infection per 100 person-days. A one-sided test shows that the P-value for this rate is 0.03, indicating significant evidence that the hospital’s infection rate is below the benchmark.

Q5. Does the change in BMI differ between the treated and placebo groups in the diet pill study?

The correct answer is Less than 0.10 but larger than 0.05.

Explanation: A two-sample t-test is used to compare the mean differences in BMI for the treated and placebo groups. Given the data and the standard deviations, the P-value for this test falls between 0.05 and 0.10, suggesting weak evidence against the null hypothesis (no difference in BMI change between the two groups).

Q6. Would you reject the null hypothesis H0:μ=1078H_0: \mu = 1078H0:μ=1078 for a 5% two-sided test given the 90% confidence interval for brain volume?

The correct answer is No you wouldn’t reject.

Explanation: Since 1078 is within the 90% confidence interval (1077 to 1123), we do not have enough evidence to reject the null hypothesis at the 5% significance level. The confidence interval suggests that the true population mean could indeed be 1078.

Q7. What is the power of the study to detect a 0.01 mm³ mean brain volume loss with a 5% one-sided test?

The correct answer is 0.70.

Explanation: The power of the study is the probability of correctly rejecting the null hypothesis when the alternative hypothesis is true. Given the sample size (100) and the standard deviation (0.04), the power of the study is approximately 0.70 for detecting a 0.01 mm³ change in brain volume.

Q8. What is the required sample size for 90% power and a 5% one-sided test to detect a 0.01 mm³ brain volume loss?

The correct answer is 160.

Explanation: The sample size needed to achieve 90% power with a 5% one-sided test for detecting a 0.01 mm³ change in brain volume is 160. This is calculated using the power analysis formula based on the desired effect size, standard deviation, and the power level.

Q9. As you increase the type one error rate, α\alphaα, what happens to power?

The correct answer is You will get larger power.

Explanation: Increasing the type one error rate (α\alphaα) increases the probability of rejecting the null hypothesis when it is false, which leads to higher statistical power. Power is inversely related to the type one error rate in hypothesis testing.

Frequently Asked Questions (FAQ)

Are the Statistical Inference quiz answers accurate?

Yes, these answers have been carefully reviewed to ensure they align with the latest course content and statistical inference principles.

Can I use these answers for both practice and graded quizzes?

Absolutely! These answers are designed for both practice quizzes and graded assessments, helping you prepare thoroughly for all evaluations.

Does this guide cover all modules of the course?

Yes, this guide provides answers for every module, ensuring complete coverage of the entire course content.

Will this guide help me improve my statistical analysis skills?

Yes, beyond providing quiz answers, this guide reinforces key concepts such as calculating confidence intervals, conducting hypothesis tests, understanding p-values, and applying statistical inference in real-world scenarios.

Conclusion

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