Statistics for Business Exam 1 Review – Step-by-Step Guidance

Q1. At α = 0.05, is the 8-week workout program effective at reducing average body fat percentage? (Paired t-test)

Background

Topic: Paired t-test for means

This question tests your ability to interpret the results of a paired t-test, which compares the means of two related groups (before and after measurements on the same subjects).

Key Terms and Formulas:

• Paired t-test: Used when comparing two related samples.

• Null hypothesis ($H_0$): No difference in means ($\mu_{before} = \mu_{after}$).

• Alternative hypothesis ($H_a$): Mean after is less than mean before ($\mu_{after} < \mu_{before}$).

• Test statistic: $t = \frac{\bar{d}}{s_d/\sqrt{n}}$

• p-value: Probability of observing a result as extreme as, or more extreme than, the observed result under $H_0$.

Step-by-Step Guidance

1. State the hypotheses: $H_0: \mu_{before} = \mu_{after}$; $H_a: \mu_{before} > \mu_{after}$ (since we are testing for reduction).

2. Check the test statistic ($t$) and the critical value for a one-tailed test at $\alpha = 0.05$ with $df = 11$.

3. Compare the calculated $t$ value to the critical $t$ value, or compare the p-value to $\alpha$.

4. If $t_{stat} > t_{critical}$ or $p < \alpha$, you have evidence to reject $H_0$.

5. Stop here and interpret what this means in the context of the problem.

Try solving on your own before revealing the answer!

Final Answer:

Since $t_{stat} = 1.81$ is greater than $t_{critical} = 1.80$ and $p = 0.049 < 0.05$, we reject the null hypothesis. There is sufficient evidence at the 0.05 level to conclude the program is effective at reducing average body fat percentage.

Q2. Did tutoring improve exam scores? (Paired t-test, mean difference = 6.4, s = 4.2, n = 10, α = 0.05)

Background

Topic: Paired t-test for means

This question tests your ability to conduct a hypothesis test for the mean difference in paired data (before and after scores for the same students).

Key Terms and Formulas:

• Paired t-test: $t = \frac{\bar{d} - \mu_0}{s_d/\sqrt{n}}$

• $\bar{d}$: Mean of the differences

• $s_d$: Standard deviation of the differences

• $n$: Number of pairs

• $\mu_0$: Hypothesized mean difference (usually 0)

Step-by-Step Guidance

1. State the hypotheses: $H_0: \mu_d = 0$ (no improvement), $H_a: \mu_d > 0$ (improvement).

2. Calculate the test statistic using $t = \frac{6.4 - 0}{4.2/\sqrt{10}}$.

3. Determine the degrees of freedom ($df = n - 1 = 9$) and find the critical $t$ value for a one-tailed test at $\alpha = 0.05$.

4. Compare your calculated $t$ value to the critical value.

5. Stop here and decide whether to reject $H_0$ based on your comparison.

Try solving on your own before revealing the answer!

Final Answer:

$t = \frac{6.4}{4.2/\sqrt{10}} \approx 4.82$. The critical value for $df=9$ at $\alpha=0.05$ (one-tailed) is about 1.833. Since $4.82 > 1.833$, we reject $H_0$. There is significant evidence that tutoring improved scores.

Q3. Is the average dinner check now greater than $28? (Population σ known, n = 49, x̄ = $30.10, σ = $6, α = 0.10)

Background

Topic: One-sample z-test for the mean (σ known)

This question tests your ability to perform a hypothesis test for the population mean when the population standard deviation is known.

Key Terms and Formulas:

• Null hypothesis: $H_0: \mu = 28$

• Alternative hypothesis: $H_a: \mu > 28$

• Test statistic: $z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}}$

• p-value: Area to the right of the calculated z in the standard normal distribution

Step-by-Step Guidance

1. State the hypotheses: $H_0: \mu = 28$, $H_a: \mu > 28$.

2. Calculate the test statistic: $z = \frac{30.10 - 28}{6/\sqrt{49}}$.

3. Find the critical z-value for $\alpha = 0.10$ (one-tailed test).

4. Compare your calculated z to the critical value, and/or find the p-value.

5. Stop here and decide whether to reject $H_0$ based on your comparison.

Try solving on your own before revealing the answer!

Final Answer:

$z = \frac{30.10 - 28}{6/\sqrt{49}} = \frac{2.10}{0.857} \approx 2.45$. The critical z for $\alpha=0.10$ is 1.28. Since $2.45 > 1.28$, we reject $H_0$. The p-value is approximately 0.007. There is sufficient evidence that the average dinner check is now greater than $28.

Q4. If x̄ = 75, s = 11, n = 60, α = 0.10, and we reject the null hypothesis, what is the probability we rejected the null when it was actually true?

Background

Topic: Type I Error (Significance Level)

This question tests your understanding of Type I error, which is the probability of rejecting the null hypothesis when it is actually true.

Key Terms and Formulas:

• Type I error: Probability of rejecting $H_0$ when $H_0$ is true.

• Significance level ($\alpha$): The probability of a Type I error.

Step-by-Step Guidance

1. Recall the definition of significance level ($\alpha$).

2. Understand that $\alpha$ is the probability of making a Type I error.

3. Since you rejected $H_0$ at $\alpha = 0.10$, the probability you made a Type I error is $\alpha$.

Try solving on your own before revealing the answer!

Final Answer:

The probability is $\alpha = 0.10$ (10%). This is the definition of the significance level.

Q5. All else equal, is it harder to reject the null in a two-tailed test or a one-tailed test?

Background

Topic: Hypothesis Testing – One-tailed vs. Two-tailed Tests

This question tests your understanding of the difference between one-tailed and two-tailed tests in terms of critical values and rejection regions.

Key Terms and Formulas:

• One-tailed test: All of $\alpha$ is in one tail.

• Two-tailed test: $\alpha$ is split between both tails.

• Critical value: The value that the test statistic must exceed to reject $H_0$.

Step-by-Step Guidance

1. Recall that in a two-tailed test, the rejection region is split between both tails, making each tail's area $\alpha/2$.

2. Compare the critical values for one-tailed and two-tailed tests at the same $\alpha$.

3. Think about which test requires a more extreme test statistic to reject $H_0$.

Try solving on your own before revealing the answer!

Final Answer:

It is harder to reject the null hypothesis in a two-tailed test because the critical value is more extreme (higher absolute value) than in a one-tailed test at the same significance level.

Q6. If σ = 6.783 and we want a 90% confidence interval with margin of error 1, what sample size should we select?

Background

Topic: Sample Size for Confidence Intervals

This question tests your ability to determine the required sample size to achieve a specified margin of error for a confidence interval when the population standard deviation is known.

Key Terms and Formulas:

• Margin of error: $E = z^* \frac{\sigma}{\sqrt{n}}$

• For a 90% confidence interval, $z^* \approx 1.645$

• Sample size formula: $n = \left( \frac{z^* \sigma}{E} \right)^2$

Step-by-Step Guidance

1. Identify the desired margin of error ($E = 1$), population standard deviation ($\sigma = 6.783$), and confidence level (90%).

2. Find the critical value $z^*$ for a 90% confidence interval ($z^* \approx 1.645$).

3. Plug the values into the sample size formula: $n = \left( \frac{z^* \sigma}{E} \right)^2$.

4. Calculate the numerator and denominator separately before squaring.

5. Stop here and complete the calculation for $n$.

Try solving on your own before revealing the answer!

Final Answer:

$n = \left( \frac{1.645 \times 6.783}{1} \right)^2 \approx (11.16)^2 \approx 124.6$. Round up to 125. You should select a sample size of 125.

Q7. How does a change in confidence level affect the width of the confidence interval?

Background

Topic: Confidence Intervals

This question tests your understanding of the relationship between confidence level and the width of a confidence interval.

Key Terms and Formulas:

• Confidence interval: $\bar{x} \pm z^* \frac{\sigma}{\sqrt{n}}$

• Confidence level: The probability that the interval contains the true parameter.

• Width: $2 \times z^* \frac{\sigma}{\sqrt{n}}$

Step-by-Step Guidance

1. Recall that a higher confidence level corresponds to a larger $z^*$ value.

2. Consider how increasing $z^*$ affects the width of the interval.

3. Think about what happens to the interval width as the confidence level increases or decreases.

Try solving on your own before revealing the answer!

Final Answer:

As the confidence level increases, the width of the confidence interval increases. This is because a higher confidence level requires a larger $z^*$ value, making the interval wider.

Q8. What is the equation for coefficient of variation and what does it measure?

Background

Topic: Measures of Relative Variation

This question tests your knowledge of the coefficient of variation, which is used to compare variability between datasets with different units or means.

Key Terms and Formulas:

• Coefficient of Variation (CV): $CV = \frac{s}{\bar{x}} \times 100\%$

• $s$: Sample standard deviation

• $\bar{x}$: Sample mean

Step-by-Step Guidance

1. Recall the formula for coefficient of variation.

2. Understand that CV expresses the standard deviation as a percentage of the mean.

3. Think about why this is useful for comparing variability across different datasets.

Try solving on your own before revealing the answer!

Final Answer:

The coefficient of variation is $CV = \frac{s}{\bar{x}} \times 100\%$. It measures the relative variability of the data, allowing comparison between datasets with different units or means.

Q9. Calculate s for the sample: 3, 7, 22, 16, 11, 9, 4, 31, 14

Background

Topic: Sample Standard Deviation

This question tests your ability to calculate the sample standard deviation from raw data.

Key Terms and Formulas:

• Sample mean: $\bar{x} = \frac{\sum x_i}{n}$

• Sample standard deviation: $s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$

Step-by-Step Guidance

1. Calculate the sample mean $\bar{x}$ by summing all values and dividing by $n$.

2. Subtract the mean from each value to find the deviations.

3. Square each deviation and sum them.

4. Divide the sum of squared deviations by $n-1$ (where $n$ is the sample size).

5. Take the square root of the result to find $s$.

Try solving on your own before revealing the answer!

Final Answer:

$s \approx 8.53$

Calculated using the formula for sample standard deviation with $n=9$.

Q10. Are SAT scores categorical data?

Background

Topic: Types of Data

This question tests your understanding of the difference between categorical and quantitative (numerical) data.

Key Terms and Formulas:

• Categorical data: Data that can be grouped into categories (e.g., colors, brands).

• Quantitative data: Data that are numerical and can be measured (e.g., scores, heights).

Step-by-Step Guidance

1. Recall the definition of categorical data.

2. Recall the definition of quantitative data.

3. Consider whether SAT scores are numbers that can be measured and averaged, or if they are categories.

Try solving on your own before revealing the answer!

Final Answer:

SAT scores are quantitative (numerical) data, not categorical data.

Q11. Calculate the regression equation for x = 11,9,13,15,12,8,16,12 and y = 13,15,17,19,22,26,28,16

Background

Topic: Simple Linear Regression

This question tests your ability to calculate the least-squares regression line for two variables.

Key Terms and Formulas:

• Regression equation: $\hat{y} = b_0 + b_1 x$

• Slope: $b_1 = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}$

• Intercept: $b_0 = \bar{y} - b_1 \bar{x}$

Step-by-Step Guidance

1. Calculate the means $\bar{x}$ and $\bar{y}$.

2. Compute $\sum (x_i - \bar{x})(y_i - \bar{y})$ and $\sum (x_i - \bar{x})^2$.

3. Calculate the slope $b_1$ using the formula above.

4. Calculate the intercept $b_0$ using the means and the slope.

5. Write the regression equation in the form $\hat{y} = b_0 + b_1 x$.

Try solving on your own before revealing the answer!

Final Answer:

The regression equation is $\hat{y} = -0.36 + 1.56x$ (rounded to two decimals).

Q12. What is the correlation coefficient for x: 7,7,7,8,9,10,11 and y: 10,10,12,13,14,15,16?

Background

Topic: Correlation Coefficient

This question tests your ability to calculate the Pearson correlation coefficient between two variables.

Key Terms and Formulas:

• Pearson correlation coefficient: $r = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum (x_i - \bar{x})^2 \sum (y_i - \bar{y})^2}}$

Step-by-Step Guidance

1. Calculate the means $\bar{x}$ and $\bar{y}$.

2. Compute the deviations $(x_i - \bar{x})$ and $(y_i - \bar{y})$ for each pair.

3. Multiply the deviations for each pair and sum them for the numerator.

4. Calculate the sum of squared deviations for $x$ and $y$ separately for the denominator.

5. Plug the values into the formula for $r$ and simplify.

Try solving on your own before revealing the answer!

Final Answer:

The correlation coefficient $r \approx 0.97$, indicating a strong positive linear relationship.

Q13. What is the probability that the mean of a sample of 33 students is more than 2.62 hours, given μ = 2.5, σ = 0.5?

Background

Topic: Sampling Distribution of the Mean (Central Limit Theorem)

This question tests your ability to find probabilities for sample means using the normal distribution.

Key Terms and Formulas:

• Sampling distribution mean: $\mu_{\bar{x}} = \mu$

• Standard error: $SE = \frac{\sigma}{\sqrt{n}}$

• Z-score: $z = \frac{\bar{x} - \mu}{SE}$

• Probability: Area to the right of the calculated z in the standard normal table

Step-by-Step Guidance

1. Calculate the standard error: $SE = \frac{0.5}{\sqrt{33}}$.

2. Find the z-score for $\bar{x} = 2.62$: $z = \frac{2.62 - 2.5}{SE}$.

3. Use the standard normal table to find the probability that $z$ is greater than your calculated value.

4. Stop here and look up the area to the right of your z-score.

Try solving on your own before revealing the answer!

Final Answer:

$SE = 0.087$, $z \approx 1.38$. The probability is about 0.084 (8.4%).

Q14. What is the probability that a sample of 38 people has a mean credit score between 690 and 720, given μ = 715, σ = 75?

Background

Topic: Sampling Distribution of the Mean

This question tests your ability to find the probability that a sample mean falls within a certain range using the normal distribution.

Key Terms and Formulas:

• Standard error: $SE = \frac{\sigma}{\sqrt{n}}$

• Z-scores: $z = \frac{\bar{x} - \mu}{SE}$

• Probability: Area between two z-scores in the standard normal table

Step-by-Step Guidance

1. Calculate the standard error: $SE = \frac{75}{\sqrt{38}}$.

2. Find the z-scores for $\bar{x} = 690$ and $\bar{x} = 720$.

3. Use the standard normal table to find the area between these two z-scores.

4. Stop here and interpret the probability.

Try solving on your own before revealing the answer!

Final Answer:

$SE = 12.16$, $z_{690} = -2.06$, $z_{720} = 0.41$. The probability is about 0.659 (65.9%).

Q15. What happens to t-scores as sample size increases?

Background

Topic: t-distribution and Sample Size

This question tests your understanding of how the t-distribution changes as sample size increases.

Key Terms and Formulas:

• t-distribution: Used when population standard deviation is unknown and sample size is small.

• Degrees of freedom: $df = n - 1$

• As $n$ increases, the t-distribution approaches the normal distribution.

Step-by-Step Guidance

1. Recall that the t-distribution is wider and has heavier tails than the normal distribution for small $n$.

2. As $n$ increases, degrees of freedom increase.

3. Think about how the critical t-values change as $n$ increases.

Try solving on your own before revealing the answer!

Final Answer:

As sample size increases, t-scores (critical values) decrease and approach the corresponding z-scores of the normal distribution.

Q16. Determine the 90% confidence interval for the mean fuel efficiency (n = 20, x̄ = 25, σ = 3)

Background

Topic: Confidence Interval for the Mean (σ known)

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

Key Terms and Formulas:

• Confidence interval: $\bar{x} \pm z^* \frac{\sigma}{\sqrt{n}}$

• For 90% confidence, $z^* \approx 1.645$

Step-by-Step Guidance

1. Identify the sample mean ($\bar{x} = 25$), population standard deviation ($\sigma = 3$), and sample size ($n = 20$).

2. Find the critical value $z^*$ for 90% confidence ($z^* = 1.645$).

3. Calculate the standard error: $SE = \frac{3}{\sqrt{20}}$.

4. Multiply $z^*$ by $SE$ to find the margin of error.

5. Stop here and set up the confidence interval formula.

Try solving on your own before revealing the answer!

Final Answer:

Margin of error $= 1.645 \times 0.671 = 1.104$. The 90% confidence interval is $25 \pm 1.10$, or (23.90, 26.10).

Q17. What type of data are histograms used to illustrate?

Background

Topic: Data Visualization

This question tests your understanding of which types of data are best represented by histograms.

Key Terms and Formulas:

• Histogram: A graphical representation of the distribution of numerical data.

• Quantitative data: Data that can be measured and ordered.

Step-by-Step Guidance

1. Recall the definition and purpose of a histogram.

2. Think about whether histograms are used for categorical or quantitative data.

3. Consider examples of data that are best illustrated with histograms.

Try solving on your own before revealing the answer!

Final Answer:

Histograms are used to illustrate the distribution of quantitative (numerical) data.

Q18. What level of measurement is a Likert scale?

Background

Topic: Levels of Measurement

This question tests your understanding of the four levels of measurement: nominal, ordinal, interval, and ratio.

Key Terms and Formulas:

• Nominal: Categories with no order

• Ordinal: Categories with a meaningful order

• Interval: Ordered, equal intervals, no true zero

• Ratio: Ordered, equal intervals, true zero

• Likert scale: Survey scale (e.g., 1 = strongly disagree to 5 = strongly agree)

Step-by-Step Guidance

1. Recall the definitions of the four levels of measurement.

2. Consider whether a Likert scale has a meaningful order and/or equal intervals.

3. Decide which level best describes a Likert scale.

Try solving on your own before revealing the answer!

Final Answer:

A Likert scale is considered ordinal data because the responses have a meaningful order but the intervals between values may not be equal.