STA1013 Chapter 9 Review – Hypothesis Testing and Statistical Inference Guidance

Q1. Formulate the null and alternative hypotheses for a hypothesis test about whether the average math SAT score at a school has changed from the 1990 mean of 506.

Background

Topic: Hypothesis Testing – Mean

This question tests your ability to correctly set up null ($H_0$) and alternative ($H_a$) hypotheses for a test about a population mean, specifically whether the mean has changed from a known value.

Key Terms:

• Null Hypothesis ($H_0$): The statement of no effect or no difference, usually includes equality.

• Alternative Hypothesis ($H_a$): The statement you are testing for, which may involve inequality.

• Population Mean ($\mu$): The average value in the population.

Step-by-Step Guidance

1. Identify the claim: The teacher wants to know if the average has changed from 506.

2. Recall that the null hypothesis ($H_0$) always includes equality: $H_0: \mu = 506$.

3. The alternative hypothesis ($H_a$) should reflect the possibility of change: $H_a: \mu \neq 506$.

4. Recognize this is a two-tailed test because the question is about any change (increase or decrease).

Try solving on your own before revealing the answer!

Q2. Formulate the null and alternative hypotheses for a test about whether the mean attendance at football games is greater than 60,100.

Background

Topic: Hypothesis Testing – Mean

This question tests your ability to set up hypotheses for a claim about a population mean being greater than a specific value.

Key Terms:

• Null Hypothesis ($H_0$): $H_0: \mu = 60,100$

• Alternative Hypothesis ($H_a$): $H_a: \mu > 60,100$

• Right-tailed test: Used when the claim is about being greater than a value.

Step-by-Step Guidance

1. Identify the claim: The owner claims the mean attendance is greater than 60,100.

2. Set the null hypothesis to include equality: $H_0: \mu = 60,100$.

3. Set the alternative hypothesis to reflect the claim: $H_a: \mu > 60,100$.

4. Recognize this is a right-tailed test.

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Q3. Formulate the null and alternative hypotheses for a test about whether the proportion of Americans who have seen a UFO is less than 0.002.

Background

Topic: Hypothesis Testing – Proportion

This question tests your ability to set up hypotheses for a claim about a population proportion.

Key Terms:

• Null Hypothesis ($H_0$): $H_0: p = 0.002$

• Alternative Hypothesis ($H_a$): $H_a: p < 0.002$

• Left-tailed test: Used when the claim is about being less than a value.

Step-by-Step Guidance

1. Identify the claim: The researcher claims the proportion is less than 0.002.

2. Set the null hypothesis to include equality: $H_0: p = 0.002$.

3. Set the alternative hypothesis to reflect the claim: $H_a: p < 0.002$.

4. Recognize this is a left-tailed test.

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Q4. State the conclusion in nontechnical terms if the null hypothesis is rejected for a cereal company's claim that the mean weight of cereal packets is at least 18 ounces.

Background

Topic: Interpreting Hypothesis Test Results

This question tests your ability to interpret the outcome of a hypothesis test in everyday language.

Key Terms:

• Null Hypothesis ($H_0$): $\mu \geq 18$ (or $\mu = 18$)

• Alternative Hypothesis ($H_a$): $\mu < 18$

• Rejecting $H_0$ means supporting $H_a$.

Step-by-Step Guidance

1. Understand what it means to reject $H_0$: There is enough evidence to support the alternative hypothesis.

2. Translate the statistical conclusion into plain language: The claim that the mean weight is at least 18 ounces is not supported.

3. Consider the direction of the test and what the evidence supports.

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Q5. State the conclusion in nontechnical terms if the null hypothesis is not rejected for a claim about the standard deviation of acetaminophen in tablets being different from 3.7 mg.

Background

Topic: Interpreting Hypothesis Test Results – Standard Deviation

This question tests your ability to interpret the outcome of a hypothesis test about variability.

Key Terms:

• Null Hypothesis ($H_0$): $\sigma = 3.7$

• Alternative Hypothesis ($H_a$): $\sigma \neq 3.7$

• Failure to reject $H_0$ means there is not enough evidence to support $H_a$.

Step-by-Step Guidance

1. Understand what it means to not reject $H_0$: There is not enough evidence to support the alternative hypothesis.

2. Translate the statistical conclusion into plain language: The claim that the standard deviation is different from 3.7 mg is not supported.

3. Consider the direction of the test and what the evidence supports.

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Q6. Find the value of the standard score, z, and determine whether the alternative hypothesis is supported for Ha: µ > 50, n = 37, x̄ = 51, σ = 3.6 at α = 0.05.

Background

Topic: Hypothesis Testing – Z-Test for Mean

This question tests your ability to calculate the z-score for a sample mean and determine if the alternative hypothesis is supported at a given significance level.

Key Formula:

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

Where:

• $\overline{x}$ = sample mean

• $\mu$ = population mean under $H_0$

• $\sigma$ = population standard deviation

• $n$ = sample size

Step-by-Step Guidance

1. Identify the known values: $\overline{x} = 51$, $\mu = 50$, $\sigma = 3.6$, $n = 37$.

2. Plug these values into the z-score formula: $z = \frac{51 - 50}{\frac{3.6}{\sqrt{37}}}$.

3. Calculate the denominator: $\frac{3.6}{\sqrt{37}}$.

4. Calculate the numerator: $51 - 50$.

5. Divide the numerator by the denominator to get the z-score.

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Q7. Find the value of the standard score, z, and determine whether the alternative hypothesis is supported for Ha: µ ≠ 18.7, n = 11, x̄ = 21, σ = 7.5 at α = 0.05.

Background

Topic: Hypothesis Testing – Z-Test for Mean (Two-Tailed)

This question tests your ability to calculate the z-score for a sample mean and determine support for a two-tailed alternative hypothesis.

Key Formula:

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

Where:

• $\overline{x}$ = sample mean

• $\mu$ = population mean under $H_0$

• $\sigma$ = population standard deviation

• $n$ = sample size

Step-by-Step Guidance

1. Identify the known values: $\overline{x} = 21$, $\mu = 18.7$, $\sigma = 7.5$, $n = 11$.

2. Plug these values into the z-score formula: $z = \frac{21 - 18.7}{\frac{7.5}{\sqrt{11}}}$.

3. Calculate the denominator: $\frac{7.5}{\sqrt{11}}$.

4. Calculate the numerator: $21 - 18.7$.

5. Divide the numerator by the denominator to get the z-score.

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Q8. Find the value of the standard score, z, and determine whether the alternative hypothesis is supported for Ha: µ < 10, n = 18, x̄ = 7.6, σ = 2.1 at α = 0.05.

Background

Topic: Hypothesis Testing – Z-Test for Mean (Left-Tailed)

This question tests your ability to calculate the z-score for a sample mean and determine support for a left-tailed alternative hypothesis.

Key Formula:

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

Where:

• $\overline{x}$ = sample mean

• $\mu$ = population mean under $H_0$

• $\sigma$ = population standard deviation

• $n$ = sample size

Step-by-Step Guidance

1. Identify the known values: $\overline{x} = 7.6$, $\mu = 10$, $\sigma = 2.1$, $n = 18$.

2. Plug these values into the z-score formula: $z = \frac{7.6 - 10}{\frac{2.1}{\sqrt{18}}}$.

3. Calculate the denominator: $\frac{2.1}{\sqrt{18}}$.

4. Calculate the numerator: $7.6 - 10$.

5. Divide the numerator by the denominator to get the z-score.

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Q9. Find the P-value for z = -1.5 for Ha: µ < 27 and determine whether the alternative hypothesis is supported at α = 0.05.

Background

Topic: P-value and Hypothesis Testing

This question tests your ability to use a z-score and a normal distribution table to find the P-value and compare it to the significance level.

Key Terms:

• P-value: The probability of observing a test statistic as extreme as, or more extreme than, the observed value under $H_0$.

• Left-tailed test: P-value is the area to the left of the z-score.

Step-by-Step Guidance

1. Locate the z-score (-1.5) in the normal distribution table.

2. Find the area to the left of z = -1.5 (this is the P-value for a left-tailed test).

3. Compare the P-value to α = 0.05 to determine support for $H_a$.

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Q10. Find the P-value for z = 2.0 for Ha: µ > 32.34 and determine whether the alternative hypothesis is supported at α = 0.05.

Background

Topic: P-value and Hypothesis Testing

This question tests your ability to use a z-score and a normal distribution table to find the P-value for a right-tailed test.

Key Terms:

• P-value: For a right-tailed test, P-value is the area to the right of the z-score.

• Right-tailed test: Used when the claim is about being greater than a value.

Step-by-Step Guidance

1. Locate the z-score (2.0) in the normal distribution table.

2. Find the area to the right of z = 2.0 (P-value = 1 - area to the left).

3. Compare the P-value to α = 0.05 to determine support for $H_a$.

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Q11. Find the P-value for z = -3.5 for Ha: µ ≠ 0.537 and determine whether the alternative hypothesis is supported at α = 0.05.

Background

Topic: P-value and Hypothesis Testing (Two-Tailed)

This question tests your ability to use a z-score and a normal distribution table to find the P-value for a two-tailed test.

Key Terms:

• P-value: For a two-tailed test, P-value = 2 × area in the tail.

• Two-tailed test: Used when the claim is about any difference.

Step-by-Step Guidance

1. Locate the z-score (-3.5) in the normal distribution table.

2. Find the area to the left of z = -3.5.

3. Multiply this area by 2 to get the P-value for a two-tailed test.

4. Compare the P-value to α = 0.05 to determine support for $H_a$.

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Q12. Conduct a full hypothesis test for the claim that tenth-graders at Birchwood High School watch less television than the city average of 22 hours. Sample: n = 32, x̄ = 19.8, σ = 7.2, α = 0.05.

Background

Topic: Hypothesis Testing – Z-Test for Mean (Left-Tailed)

This question tests your ability to conduct a hypothesis test for a population mean and interpret the result.

Key Formula:

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

Step-by-Step Guidance

1. State the hypotheses: $H_0: \mu = 22$, $H_a: \mu < 22$.

2. Identify the known values: $\overline{x} = 19.8$, $\mu = 22$, $\sigma = 7.2$, $n = 32$.

3. Plug these values into the z-score formula: $z = \frac{19.8 - 22}{\frac{7.2}{\sqrt{32}}}$.

4. Calculate the denominator: $\frac{7.2}{\sqrt{32}}$.

5. Calculate the numerator: $19.8 - 22$.

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Q13. Conduct a full hypothesis test for the claim that the average math SAT score at a school has dropped from 475. Sample: n = 70, x̄ = 469, σ = 73, α = 0.05.

Background

Topic: Hypothesis Testing – Z-Test for Mean (Left-Tailed)

This question tests your ability to conduct a hypothesis test for a population mean and interpret the result.

Key Formula:

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

Step-by-Step Guidance

1. State the hypotheses: $H_0: \mu = 475$, $H_a: \mu < 475$.

2. Identify the known values: $\overline{x} = 469$, $\mu = 475$, $\sigma = 73$, $n = 70$.

3. Plug these values into the z-score formula: $z = \frac{469 - 475}{\frac{73}{\sqrt{70}}}$.

4. Calculate the denominator: $\frac{73}{\sqrt{70}}$.

5. Calculate the numerator: $469 - 475$.

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Q14. Identify the Type I error for a test about whether the mean running time of flashlight batteries has increased from 9.0 hours.

Background

Topic: Errors in Hypothesis Testing

This question tests your understanding of Type I error, which is rejecting a true null hypothesis.

Key Terms:

• Type I error: Rejecting $H_0$ when it is actually true.

• Type II error: Failing to reject $H_0$ when it is actually false.

Step-by-Step Guidance

1. Recall the definition of Type I error.

2. Apply it to the context: Rejecting the claim that the mean is 9.0 hours when it actually is 9.0 hours.

3. Identify the answer choice that matches this definition.

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Q15. Identify the Type II error for a test about whether the mean running time of flashlight batteries has increased from 9.4 hours.

Background

Topic: Errors in Hypothesis Testing

This question tests your understanding of Type II error, which is failing to reject a false null hypothesis.

Key Terms:

• Type I error: Rejecting $H_0$ when it is actually true.

• Type II error: Failing to reject $H_0$ when it is actually false.

Step-by-Step Guidance

1. Recall the definition of Type II error.

2. Apply it to the context: Failing to reject the claim that the mean is 9.4 hours when it is actually greater than 9.4 hours.

3. Identify the answer choice that matches this definition.

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Q16. Find the P-value for a test of the claim that more than 28% of medical school students plan to go into general practice. Sample: n = 130, p̂ = 0.32.

Background

Topic: Hypothesis Testing – Proportion (Right-Tailed)

This question tests your ability to calculate the z-score for a sample proportion and find the P-value for a right-tailed test.

Key Formula:

$z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

Where:

• $\hat{p}$ = sample proportion

• $p$ = population proportion under $H_0$

• $n$ = sample size

Step-by-Step Guidance

1. State the hypotheses: $H_0: p = 0.28$, $H_a: p > 0.28$.

2. Calculate the z-score: $z = \frac{0.32 - 0.28}{\sqrt{\frac{0.28 \times 0.72}{130}}}$.

3. Find the area to the right of the calculated z-score (P-value).

4. Compare the P-value to α = 0.05 to determine support for $H_a$.

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Q17. Find the P-value for a test of the claim that fewer than 6% of fax machines are defective. Sample: n = 97, p̂ = 0.05.

Background

Topic: Hypothesis Testing – Proportion (Left-Tailed)

This question tests your ability to calculate the z-score for a sample proportion and find the P-value for a left-tailed test.

Key Formula:

$z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

Where:

• $\hat{p}$ = sample proportion

• $p$ = population proportion under $H_0$

• $n$ = sample size

Step-by-Step Guidance

1. State the hypotheses: $H_0: p = 0.06$, $H_a: p < 0.06$.

2. Calculate the z-score: $z = \frac{0.05 - 0.06}{\sqrt{\frac{0.06 \times 0.94}{97}}}$.

3. Find the area to the left of the calculated z-score (P-value).

4. Compare the P-value to α = 0.05 to determine support for $H_a$.

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Q18. Find the P-value for a test of the claim that the percentage of forty-year-old men who smoke is 22%. Sample: n = 139, p̂ = 0.26.

Background

Topic: Hypothesis Testing – Proportion (Two-Tailed)

This question tests your ability to calculate the z-score for a sample proportion and find the P-value for a two-tailed test.

Key Formula:

$z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

Where:

• $\hat{p}$ = sample proportion

• $p$ = population proportion under $H_0$

• $n$ = sample size

Step-by-Step Guidance

1. State the hypotheses: $H_0: p = 0.22$, $H_a: p \neq 0.22$.

2. Calculate the z-score: $z = \frac{0.26 - 0.22}{\sqrt{\frac{0.22 \times 0.78}{139}}}$.

3. Find the area in both tails for the calculated z-score (P-value = 2 × area in tail).

4. Compare the P-value to α = 0.05 to determine support for $H_a$.

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Q19. Conduct a full hypothesis test for the claim that no more than 1% of 3.5" disks are defective. Sample: n = 800, p̂ = 0.03, α = 0.05.

Background

Topic: Hypothesis Testing – Proportion (Right-Tailed)

This question tests your ability to conduct a hypothesis test for a population proportion and interpret the result.

Key Formula:

$z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

Step-by-Step Guidance

1. State the hypotheses: $H_0: p = 0.01$, $H_a: p > 0.01$.

2. Calculate the z-score: $z = \frac{0.03 - 0.01}{\sqrt{\frac{0.01 \times 0.99}{800}}}$.

3. Find the area to the right of the calculated z-score (P-value).

4. Compare the P-value to α = 0.05 to determine support for $H_a$.

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Q20. Conduct a full hypothesis test for the claim that the proportion of adults exposed to a flu strain is different from 8%. Sample: n = 47, x = 8, α = 0.05.

Background

Topic: Hypothesis Testing – Proportion (Two-Tailed)

This question tests your ability to conduct a hypothesis test for a population proportion and interpret the result.

Key Formula:

$\hat{p} = \frac{x}{n}$

$z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

Step-by-Step Guidance

1. Calculate the sample proportion: $\hat{p} = \frac{8}{47}$.

2. State the hypotheses: $H_0: p = 0.08$, $H_a: p \neq 0.08$.

3. Calculate the z-score: $z = \frac{\hat{p} - 0.08}{\sqrt{\frac{0.08 \times 0.92}{47}}}$.

4. Find the area in both tails for the calculated z-score (P-value = 2 × area in tail).

5. Compare the P-value to α = 0.05 to determine support for $H_a$.

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