Step-by-Step Guidance for Inference Practice Problems (Statistics)

Q1. What is the probability that the sample mean spending is greater than $55, given a population mean of $52, population standard deviation of $12, and a sample size of 64?

Background

Topic: Sampling Distributions and Central Limit Theorem (CLT)

This question tests your understanding of the sampling distribution of the sample mean and how to calculate probabilities using the normal distribution.

Key Terms and Formulas

• Population mean ($\mu$): the average of the entire population.

• Population standard deviation ($\sigma$): the standard deviation of the entire population.

• Sample size ($n$): the number of observations in the sample.

• Standard error of the mean ($SE$): $SE = \frac{\sigma}{\sqrt{n}}$

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

Step-by-Step Guidance

1. Identify the known values: $\mu = 52$, $\sigma = 12$, $n = 64$, and the sample mean of interest $\bar{x} = 55$.

2. Calculate the standard error of the mean using $SE = \frac{\sigma}{\sqrt{n}}$.

3. Compute the Z-score for $\bar{x} = 55$ using $Z = \frac{\bar{x} - \mu}{SE}$.

4. Use the standard normal distribution to find the probability that the sample mean is greater than $55$(i.e.,$P(\bar{x} > 55)$), which is the area to the right of your calculated Z-score.

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Q2. What is the probability that more than 25 products are returned, given 18% return rate and a sample of 150 products?

Background

Topic: Sampling Distribution of a Proportion (Binomial to Normal Approximation)

This question tests your ability to use the normal approximation to the binomial distribution to find probabilities about sample counts or proportions.

Key Terms and Formulas

• Population proportion ($p$): the probability of a product being returned ($p = 0.18$).

• Sample size ($n$): number of products sampled ($n = 150$).

• Sample count ($X$): number of returned products.

• Mean of $X$: $\mu_X = np$

• Standard deviation of $X$: $\sigma_X = \sqrt{np(1-p)}$

• Z-score: $Z = \frac{X - \mu_X}{\sigma_X}$

Step-by-Step Guidance

1. Identify the known values: $p = 0.18$, $n = 150$, and the count of interest $X = 25$.

2. Calculate the mean and standard deviation: $\mu_X = np$, $\sigma_X = \sqrt{np(1-p)}$.

3. Apply the continuity correction: since you want $P(X > 25)$, use $P(X \geq 26)$, so use $X = 25.5$ for the normal approximation.

4. Compute the Z-score: $Z = \frac{25.5 - \mu_X}{\sigma_X}$.

5. Find the probability to the right of this Z-score using the standard normal table.

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Q3. What is the probability that the sample mean delivery time is between 38 and 42 minutes, given a normal distribution with mean 40 and standard deviation 8, and a sample of 36 deliveries?

Background

Topic: Sampling Distribution of the Mean (Normal Distribution)

This question tests your ability to use the sampling distribution of the mean to find probabilities for a sample mean from a normal population.

Key Terms and Formulas

• Population mean ($\mu$): 40 minutes

• Population standard deviation ($\sigma$): 8 minutes

• Sample size ($n$): 36

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

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

Step-by-Step Guidance

1. Identify the known values: $\mu = 40$, $\sigma = 8$, $n = 36$, and the interval $38 < \bar{x} < 42$.

2. Calculate the standard error: $SE = \frac{8}{\sqrt{36}}$.

3. Compute the Z-scores for $\bar{x} = 38$ and $\bar{x} = 42$.

4. Use the standard normal table to find the probabilities corresponding to each Z-score, then find the probability between them.

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Q4. What is the probability that the sample proportion is less than 0.58, given 62% of customers use online banking and a sample of 200?

Background

Topic: Sampling Distribution of a Proportion

This question tests your ability to use the normal approximation for the sampling distribution of a sample proportion.

Key Terms and Formulas

• Population proportion ($p$): 0.62

• Sample size ($n$): 200

• Sample proportion ($\hat{p}$): value of interest is 0.58

• Standard error: $SE = \sqrt{\frac{p(1-p)}{n}}$

• Z-score: $Z = \frac{\hat{p} - p}{SE}$

Step-by-Step Guidance

1. Identify the known values: $p = 0.62$, $n = 200$, $\hat{p} = 0.58$.

2. Calculate the standard error: $SE = \sqrt{\frac{0.62 \times 0.38}{200}}$.

3. Compute the Z-score: $Z = \frac{0.58 - 0.62}{SE}$.

4. Use the standard normal table to find the probability to the left of this Z-score.

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Q5. Construct a 95% confidence interval for the mean purchase, given a sample of 49 transactions with mean $84 and sample standard deviation $14.

Background

Topic: Confidence Interval for a Mean (t-distribution)

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

Key Terms and Formulas

• Sample mean ($\bar{x}$): 84

• Sample standard deviation ($s$): 14

• Sample size ($n$): 49

• Degrees of freedom ($df$): $n-1$

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

• Confidence interval: $\bar{x} \pm t^* \times SE$

Step-by-Step Guidance

1. Identify the known values: $\bar{x} = 84$, $s = 14$, $n = 49$.

2. Calculate the standard error: $SE = \frac{14}{\sqrt{49}}$.

3. Determine the degrees of freedom: $df = 49 - 1$.

4. Find the critical t-value ($t^*$) for a 95% confidence interval with $df$ degrees of freedom.

5. Set up the confidence interval formula: $84 \pm t^* \times SE$.

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Q6. Construct a 90% confidence interval for the proportion of satisfied customers, given 198 out of 300 were satisfied.

Background

Topic: Confidence Interval for a Proportion

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

Key Terms and Formulas

• Sample proportion ($\hat{p}$): $\hat{p} = \frac{198}{300}$

• Sample size ($n$): 300

• Standard error: $SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

• Critical z-value ($z^*$) for 90% confidence

• Confidence interval: $\hat{p} \pm z^* \times SE$

Step-by-Step Guidance

1. Calculate the sample proportion: $\hat{p} = \frac{198}{300}$.

2. Compute the standard error: $SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{300}}$.

3. Find the critical z-value ($z^*$) for a 90% confidence interval.

4. Set up the confidence interval: $\hat{p} \pm z^* \times SE$.

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Q7. Construct a 99% confidence interval for the mean hours worked per week, given a sample of 36 employees with mean 41 and standard deviation 6.

Background

Topic: Confidence Interval for a Mean (t-distribution)

This question tests your ability to construct a confidence interval for a population mean using the t-distribution.

Key Terms and Formulas

• Sample mean ($\bar{x}$): 41

• Sample standard deviation ($s$): 6

• Sample size ($n$): 36

• Degrees of freedom ($df$): $n-1$

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

• Confidence interval: $\bar{x} \pm t^* \times SE$

Step-by-Step Guidance

1. Identify the known values: $\bar{x} = 41$, $s = 6$, $n = 36$.

2. Calculate the standard error: $SE = \frac{6}{\sqrt{36}}$.

3. Determine the degrees of freedom: $df = 36 - 1$.

4. Find the critical t-value ($t^*$) for a 99% confidence interval with $df$ degrees of freedom.

5. Set up the confidence interval: $41 \pm t^* \times SE$.

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Q8. Test the claim that the average bottle contains 500 mL, given a sample of 40 bottles with mean 492 mL and standard deviation 20 mL, at the 5% significance level.

Background

Topic: Hypothesis Testing for a Mean (One-Sample t-test)

This question tests your ability to perform a hypothesis test for a population mean using sample data.

Key Terms and Formulas

• Null hypothesis ($H_0$): $\mu = 500$

• Alternative hypothesis ($H_a$): $\mu \neq 500$

• Sample mean ($\bar{x}$): 492

• Sample standard deviation ($s$): 20

• Sample size ($n$): 40

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

• Degrees of freedom: $n-1$

Step-by-Step Guidance

1. State the null and alternative hypotheses.

2. Calculate the standard error: $SE = \frac{20}{\sqrt{40}}$.

3. Compute the test statistic: $t = \frac{492 - 500}{SE}$.

4. Determine the critical t-value for a two-tailed test at $\alpha = 0.05$ with $n-1$ degrees of freedom.

5. Compare the calculated t-value to the critical value to decide whether to reject the null hypothesis.

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Q9. Test the claim that at least 70% of customers renew their subscriptions, given 78 out of 120 renewed, at α = 0.10.

Background

Topic: Hypothesis Testing for a Proportion (One-Sample z-test)

This question tests your ability to perform a hypothesis test for a population proportion.

Key Terms and Formulas

• Null hypothesis ($H_0$): $p \geq 0.70$

• Alternative hypothesis ($H_a$): $p < 0.70$

• Sample proportion ($\hat{p}$): $\hat{p} = \frac{78}{120}$

• Standard error: $SE = \sqrt{\frac{p_0(1-p_0)}{n}}$

• Test statistic: $z = \frac{\hat{p} - p_0}{SE}$

Step-by-Step Guidance

1. State the null and alternative hypotheses.

2. Calculate the sample proportion: $\hat{p} = \frac{78}{120}$.

3. Compute the standard error: $SE = \sqrt{\frac{0.70 \times 0.30}{120}}$.

4. Calculate the test statistic: $z = \frac{\hat{p} - 0.70}{SE}$.

5. Compare the calculated z-value to the critical value for $\alpha = 0.10$ (one-tailed test).

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Q10. Construct a 95% confidence interval for the difference in mean daily sales between Store A and Store B, given their sample sizes, means, and standard deviations.

Background

Topic: Confidence Interval for the Difference Between Two Means (Independent Samples)

This question tests your ability to construct a confidence interval for the difference between two independent means.

Key Terms and Formulas

• Sample means: $\bar{x}_A = 5200$, $\bar{x}_B = 4900$

• Sample standard deviations: $s_A = 600$, $s_B = 700$

• Sample sizes: $n_A = 40$, $n_B = 35$

• Standard error: $SE = \sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}$

• Confidence interval: $(\bar{x}_A - \bar{x}_B) \pm t^* \times SE$

Step-by-Step Guidance

1. Identify the known values for both stores.

2. Calculate the standard error: $SE = \sqrt{\frac{600^2}{40} + \frac{700^2}{35}}$.

3. Determine the degrees of freedom (use the formula for unequal variances if needed).

4. Find the critical t-value ($t^*$) for a 95% confidence interval with the calculated degrees of freedom.

5. Set up the confidence interval: $(5200 - 4900) \pm t^* \times SE$.

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Q11. Construct a 90% confidence interval for the difference in proportions of customers who purchased between two advertising campaigns (Online: 48/200, TV: 30/180).

Background

Topic: Confidence Interval for the Difference Between Two Proportions

This question tests your ability to construct a confidence interval for the difference between two independent proportions.

Key Terms and Formulas

• Sample proportions: $\hat{p}_1 = \frac{48}{200}$, $\hat{p}_2 = \frac{30}{180}$

• Standard error: $SE = \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}$

• Confidence interval: $(\hat{p}_1 - \hat{p}_2) \pm z^* \times SE$

Step-by-Step Guidance

1. Calculate the sample proportions for both campaigns.

2. Compute the standard error for the difference in proportions.

3. Find the critical z-value ($z^*$) for a 90% confidence interval.

4. Set up the confidence interval: $(\hat{p}_1 - \hat{p}_2) \pm z^* \times SE$.

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Q12. Test whether the mean lifetimes of batteries from two factories are different, given their sample sizes, means, and standard deviations, at α = 0.05.

Background

Topic: Hypothesis Testing for the Difference Between Two Means (Independent Samples t-test)

This question tests your ability to perform a two-sample t-test for the difference between two means.

Key Terms and Formulas

• Null hypothesis ($H_0$): $\mu_A = \mu_B$

• Alternative hypothesis ($H_a$): $\mu_A \neq \mu_B$

• Sample means: $\bar{x}_A = 420$, $\bar{x}_B = 395$

• Sample standard deviations: $s_A = 40$, $s_B = 50$

• Sample sizes: $n_A = 30$, $n_B = 35$

• Standard error: $SE = \sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}$

• Test statistic: $t = \frac{\bar{x}_A - \bar{x}_B}{SE}$

• Degrees of freedom: use the formula for unequal variances if needed.

Step-by-Step Guidance

1. State the null and alternative hypotheses.

2. Calculate the standard error for the difference in means.

3. Compute the test statistic: $t = \frac{420 - 395}{SE}$.

4. Determine the degrees of freedom and the critical t-value for $\alpha = 0.05$ (two-tailed test).

5. Compare the calculated t-value to the critical value to decide whether to reject the null hypothesis.

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Q13. Test whether the new website has a higher purchase rate than the old website, given the number of visitors and purchases for each, at α = 0.05.

Background

Topic: Hypothesis Testing for the Difference Between Two Proportions (One-Tailed Test)

This question tests your ability to perform a hypothesis test for the difference between two proportions.

Key Terms and Formulas

• Null hypothesis ($H_0$): $p_{new} \leq p_{old}$

• Alternative hypothesis ($H_a$): $p_{new} > p_{old}$

• Sample proportions: $\hat{p}_{old} = \frac{40}{250}$, $\hat{p}_{new} = \frac{66}{300}$

• Pooled proportion: $\hat{p} = \frac{40 + 66}{250 + 300}$

• Standard error: $SE = \sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_{old}} + \frac{1}{n_{new}}\right)}$

• Test statistic: $z = \frac{\hat{p}_{new} - \hat{p}_{old}}{SE}$

Step-by-Step Guidance

1. State the null and alternative hypotheses.

2. Calculate the sample proportions for both websites.

3. Compute the pooled proportion.

4. Calculate the standard error using the pooled proportion.

5. Compute the test statistic: $z = \frac{\hat{p}_{new} - \hat{p}_{old}}{SE}$.

6. Compare the calculated z-value to the critical value for $\alpha = 0.05$ (one-tailed test).

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