Comprehensive Statistics Final Exam Study Guide

Q1. A survey of 250 students at a large university found that 70% owned a laptop computer. What is the population? What is the sample?

Background

Topic: Population vs. Sample

This question tests your understanding of the difference between a population and a sample in statistics.

Key Terms:

• Population: The entire group of individuals or items that you want to study.

• Sample: A subset of the population that is actually observed or surveyed.

Step-by-Step Guidance

1. Identify what group the survey is trying to learn about (all students at the university).

2. Determine how many students were actually surveyed (250 students).

3. Think about which group represents the population and which represents the sample.

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Q2. Suzanne owns a diner. Last week, she counted 230 people who walked by her diner, 150 of whom came into the diner. Of the 150, only 45 ordered dessert. Find the probability that a person who walks past the diner will actually enter the diner.

Background

Topic: Basic Probability

This question tests your ability to calculate probability based on observed frequencies.

Key Formula:

$\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

Step-by-Step Guidance

1. Identify the total number of people who walked by the diner (230).

2. Identify the number of people who entered the diner (150).

3. Set up the probability formula using these numbers.

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Q3. Suppose there are 8 white, 4 black, 12 red, and 6 blue beads in a bag. Two beads are taken from the bag (with or without replacement). Find various probabilities.

Background

Topic: Probability with and without replacement

This question tests your understanding of probability calculations for events with and without replacement.

Key Terms and Formulas:

• With replacement: Each draw is independent; the total number of beads stays the same.

• Without replacement: Each draw is dependent; the total number of beads decreases after each draw.

• Probability of two events: $P(A \text{ and } B) = P(A) \times P(B)$ (if independent)

Step-by-Step Guidance (for part a: probability both are red, with replacement)

1. Find the total number of beads: $8 + 4 + 12 + 6$.

2. Calculate the probability of drawing a red bead on the first draw: $\frac{12}{\text{total}}$.

3. Since it's with replacement, the probability of drawing a red bead on the second draw is also $\frac{12}{\text{total}}$.

4. Multiply the probabilities for both draws.

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Q4. Use the contingency table to answer probability questions about age and text messaging plans.

Background

Topic: Probability from Contingency Tables

This question tests your ability to use a contingency table to calculate probabilities, including conditional probabilities.

Key Terms and Formulas:

• Marginal probability: Probability of a single event occurring.

• Conditional probability: Probability of an event given another event has occurred.

• $P(A|B) = \frac{P(A \text{ and } B)}{P(B)}$

Step-by-Step Guidance (for part a: probability of limited text plan)

1. Find the total number of people surveyed (sum of all entries).

2. Find the total number of people with a limited text plan (sum of the 'limited text plan' column).

3. Set up the probability formula: $\frac{\text{number with limited plan}}{\text{total surveyed}}$.

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Q5. A local survey found that 24% of drivers do not wear a seat belt at all times. In 20 traffic stops, find various binomial probabilities.

Background

Topic: Binomial Probability Distribution

This question tests your ability to use the binomial probability formula to calculate probabilities for a fixed number of trials.

Key Formula:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

Where:

• $n$ = number of trials (traffic stops)

• $k$ = number of successes (drivers not wearing seat belt)

• $p$ = probability of success (0.24)

Step-by-Step Guidance (for part b: exactly 7 not wearing seat belt)

1. Identify $n = 20$, $k = 7$, $p = 0.24$.

2. Plug these values into the binomial formula.

3. Calculate $\binom{20}{7}$, $0.24^7$, and $0.76^{13}$ separately.

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Q6. A recent survey found that 64% of school children bring their own lunch to school. If 25 children are randomly selected, find various binomial probabilities.

Background

Topic: Binomial Probability Distribution

This question tests your ability to calculate binomial probabilities for a given number of trials and probability of success.

Key Formula:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

Where:

• $n$ = 25

• $p$ = 0.64

Step-by-Step Guidance (for part a: at least 18 bring their own lunch)

1. Identify the values: $n = 25$, $p = 0.64$.

2. For 'at least 18', calculate $P(X \geq 18)$, which means summing probabilities for $X = 18$ to $X = 25$.

3. Set up the binomial formula for each value of $k$ from 18 to 25.

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Q7. Find the following probabilities for the standard normal distribution:

Background

Topic: Standard Normal Distribution

This question tests your ability to use the standard normal table (z-table) to find probabilities associated with z-scores.

Key Terms:

• Standard normal distribution: A normal distribution with mean 0 and standard deviation 1.

• Z-score: Number of standard deviations from the mean.

Step-by-Step Guidance (for part a: $P(z < 1.53)$)

1. Identify the z-score: $z = 1.53$.

2. Use the z-table to find the area to the left of $z = 1.53$.

3. Interpret the value as the probability that a randomly selected value is less than 1.53 standard deviations above the mean.

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Q8. Find the z-score given a certain area to the right or left.

Background

Topic: Inverse Normal Calculations

This question tests your ability to use the z-table in reverse to find the z-score corresponding to a given probability.

Key Terms:

• Area to the right: $P(z > z_0)$

• Area to the left: $P(z < z_0)$

Step-by-Step Guidance (for part a: 26% area to the right of z)

1. Recognize that the area to the right is 0.26, so the area to the left is 0.74.

2. Use the z-table to find the z-score where the area to the left is 0.74.

3. Interpret the z-score as the value where 26% of the distribution lies above it.

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Q9. The number of hours of TV watched per day by children ages 7-14 is normally distributed with a mean of 4.2 hours and a standard deviation of 1.8 hours. What is the probability that a child selected at random watches less than 3.9 hours of TV per day?

Background

Topic: Normal Distribution Probability

This question tests your ability to calculate probabilities for a normal distribution using z-scores.

Key Formula:

$z = \frac{x - \mu}{\sigma}$

Where:

• $x$ = value of interest (hours watched)

• $\mu$ = mean (4.2)

• $\sigma$ = standard deviation (1.8)

Step-by-Step Guidance

1. Identify the values: $x = 3.9$, $\mu = 4.2$, $\sigma = 1.8$.

2. Calculate the z-score using the formula.

3. Use the z-table to find the probability corresponding to this z-score.

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Q10. The scores on a recent statistics exam were normally distributed with a mean of 84.5 and a standard deviation of 6. The statistics honor society requires that a student must be in the top 5% of his/her class to be accepted. What is the minimum test score a student must have to be accepted?

Background

Topic: Normal Distribution Percentiles

This question tests your ability to find a value corresponding to a percentile in a normal distribution.

Key Formula:

$z = \frac{x - \mu}{\sigma}$

Where:

• $x$ = score to find

• $\mu$ = mean (84.5)

• $\sigma$ = standard deviation (6)

Step-by-Step Guidance

1. Recognize that the top 5% corresponds to the 95th percentile.

2. Use the z-table to find the z-score corresponding to the 95th percentile.

3. Set up the formula $x = \mu + z \sigma$ to solve for the minimum score.

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Q11. The ages of students involved in extra-curricular activities are normally distributed with a mean of 21.6 years and a standard deviation of 3.4 years. If a sample of 38 applicants is randomly selected, find the probability that they will have a mean age greater than 22 years.

Background

Topic: Sampling Distribution of the Mean

This question tests your ability to use the central limit theorem and normal distribution to find probabilities about sample means.

Key Formula:

$z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}$

Where:

• $\bar{x}$ = sample mean (22)

• $\mu$ = population mean (21.6)

• $\sigma$ = population standard deviation (3.4)

• $n$ = sample size (38)

Step-by-Step Guidance

1. Calculate the standard error: $\frac{\sigma}{\sqrt{n}}$.

2. Calculate the z-score for $\bar{x} = 22$.

3. Use the z-table to find the probability that the sample mean is greater than 22.

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Q12. Given the confidence interval, 3.54 < μ < 7.68, and that σ = 1.24 and n = 85, find the margin of error, E.

Background

Topic: Confidence Intervals

This question tests your ability to interpret a confidence interval and calculate the margin of error.

Key Formula:

$E = \frac{\text{Upper limit} - \text{Lower limit}}{2}$

Step-by-Step Guidance

1. Identify the upper and lower limits of the confidence interval (7.68 and 3.54).

2. Subtract the lower limit from the upper limit.

3. Divide the result by 2 to find the margin of error.

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Q13. A random sample of 26 test grades has a mean of 84.7 and a standard deviation of 3.6. Compute a 95% confidence interval for the mean test grade of the class.

Background

Topic: Confidence Intervals for the Mean (t-distribution)

This question tests your ability to construct a confidence interval for the mean when the sample size is small and the population standard deviation is unknown.

Key Formula:

$\bar{x} \pm t_{\alpha/2} \frac{s}{\sqrt{n}}$

Where:

• $\bar{x}$ = sample mean (84.7)

• $s$ = sample standard deviation (3.6)

• $n$ = sample size (26)

• $t_{\alpha/2}$ = t-score for 95% confidence and 25 degrees of freedom

Step-by-Step Guidance

1. Calculate the standard error: $\frac{s}{\sqrt{n}}$.

2. Find the appropriate t-score for 95% confidence and 25 degrees of freedom.

3. Multiply the t-score by the standard error to find the margin of error.

4. Add and subtract the margin of error from the sample mean to get the confidence interval.

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Q14. A random sample of 18 border collies has a mean of 46 pounds. A previous study indicates that the standard deviation of the weight of all border collies is 7.8 pounds. Compute a 99% confidence interval of the weight of all border collies.

Background

Topic: Confidence Intervals for the Mean (z-distribution)

This question tests your ability to construct a confidence interval for the mean when the population standard deviation is known.

Key Formula:

$\bar{x} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

Where:

• $\bar{x}$ = sample mean (46)

• $\sigma$ = population standard deviation (7.8)

• $n$ = sample size (18)

• $z_{\alpha/2}$ = z-score for 99% confidence

Step-by-Step Guidance

1. Calculate the standard error: $\frac{\sigma}{\sqrt{n}}$.

2. Find the z-score for 99% confidence.

3. Multiply the z-score by the standard error to find the margin of error.

4. Add and subtract the margin of error from the sample mean to get the confidence interval.

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Q15. In a recent survey of 514 SUV owners, 154 said they were satisfied with their gas mileage. Compute a 90% confidence interval for the proportion of SUV owners who say they are satisfied with their gas mileage.

Background

Topic: Confidence Interval for a Proportion

This question tests your ability to construct a confidence interval for a population proportion.

Key Formula:

$\hat{p} \pm z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

Where:

• $\hat{p}$ = sample proportion ($\frac{154}{514}$)

• $n$ = sample size (514)

• $z_{\alpha/2}$ = z-score for 90% confidence

Step-by-Step Guidance

1. Calculate the sample proportion: $\hat{p} = \frac{154}{514}$.

2. Calculate the standard error: $\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$.

3. Find the z-score for 90% confidence.

4. Multiply the z-score by the standard error to find the margin of error.

5. Add and subtract the margin of error from the sample proportion to get the confidence interval.

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Q16. Assume that we want to estimate the mean SAT score of a class of incoming freshmen. How many students must be included in the sample if we want to be 95% confident that our estimate will be within 4 points of the population mean? Assume the population standard deviation is known to be 20.

Background

Topic: Sample Size Calculation for Estimating a Mean

This question tests your ability to use the sample size formula for estimating a mean with a specified margin of error.

Key Formula:

$n = \left( \frac{z_{\alpha/2} \sigma}{E} \right)^2$

Where:

• $z_{\alpha/2}$ = z-score for 95% confidence

• $\sigma$ = population standard deviation (20)

• $E$ = margin of error (4)

Step-by-Step Guidance

1. Identify the values: $z_{\alpha/2}$ for 95% confidence, $\sigma = 20$, $E = 4$.

2. Plug these values into the sample size formula.

3. Calculate the numerator and denominator separately before squaring the result.

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Q17. The following table displays the fat grams and the number of calories in fast food hamburgers. Construct a scatter plot, calculate the correlation coefficient, find the least squares regression equation, and interpret results.

Background

Topic: Correlation and Regression

This question tests your ability to analyze bivariate data, calculate correlation, and perform linear regression.

Key Terms and Formulas:

• Scatter plot: Visual representation of the relationship between two variables.

• Correlation coefficient ($r$): Measures the strength and direction of linear relationship.

• Least squares regression equation: $y = a + bx$

• Coefficient of determination ($r^2$): Proportion of variance explained by the model.

Step-by-Step Guidance (for part b: calculate the correlation coefficient)

1. List the pairs of fat grams and calories.

2. Calculate the mean of fat grams and the mean of calories.

3. Use the formula for $r$:

   $r = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum (x_i - \bar{x})^2 \sum (y_i - \bar{y})^2}}$

4. Compute the numerator and denominator separately.

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