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Practice Problems for Statistics Exam 3
Problem Set 1: Population Proportions and Sample Proportions
- 1. Distribution of Adult Smokers
- National data indicates that 44% of the adult population had never smoked.
- In a random sample of 100 adults, 30 had never smoked.
- Determine which statement is true:
- (a) $p = 0.44$, $ar{p} = 0.30$, and $ ext{SE}{ar{p}} = 0.05$ - (b) $p = 0.44$, $ar{p} = 0.30$, and $ ext{SE}{ar{p}} = 0.46$
- (c) $p = 0.30$, $ar{p} = 0.44$, and $ ext{SE}{ar{p}} = 0.23$ - (d) $p = 0.30$, $ar{p} = 0.44$, and $ ext{SE}{ar{p}} = 0.05$
Problem Set 2: Central Limit Theorem and Sample Sizes
2. Statistics Requirement at NIU
- 60% of NIU students require statistics.
- A sample of 100 students yields a sample proportion.
- The sample follows a Normal Model with:
- Mean: $ar{p} = 0.60$
- Standard Deviation: $ ext{SE}_{ar{p}} = 0.05$
- Probability that the sample proportion exceeds 0.53 is:
- (a) 0.0808
- (b) 0.9192
- (c) 0.8186
- (d) 0.84003. Central Limit Theorem Summary
- The best summary of the Central Limit Theorem:
- (a) The sampling distribution of the sample mean will be approximately Normal if the sample size $n$ is large.
- (b) The distribution of the population of interest will always be Normal.
- (c) The distribution of the population of interest will be approximately Normal if the sample size $n$ is large.
- (d) The sampling distribution of the sample mean will always be normal regardless of the shape of the population.4. Cars Per Household in DeKalb
- Average number of cars per household: 1.2
- Standard deviation: 1.5
- Random sample of 100 households analyzed:
- Sampling distribution of $ar{x}$:
- (a) approximately normal with mean $ar{x} = 1.2$, SD $ ext{SE}{ar{x}} = 1.5$. - (b) approximately normal with mean $ar{x} = 1.2$, SD $ ext{SE}{ar{x}} = 0.15$.
- (c) approximately normal with mean $ar{x} = 1.2$, SD $ ext{SE}_{ar{x}} = 0.015$.
- (d) not approximately normal because the population distribution is not normal.
Problem Set 3: Sample Means and Distributions
5. Sample Statistics from Car Battery Lifetimes
- Mean lifetime: 48 months
- Standard deviation: 6 months
- If a sample of 100 is selected, determine:
- (a) mean = 48, $ ext{SE}{ar{x}} = 6$ - (b) mean = 48, $ ext{SE}{ar{x}} = 0.6$
- (c) mean = 4.8, $ ext{SE}{ar{x}} = 6$ - (d) mean = 4.8, $ ext{SE}{ar{x}} = 0.6$6. Distribution of Sample Means with Sample Size n = 100
- For a random sample of size 100, the sampling distribution:
- (a) Approximately normal
- (b) Right skewed
- (c) Left skewed
- (d) None of the above7. Probability of Sample Mean Less than 47
- Find probability that the sample mean for the sample of 100 is less than 47 months:
- (a) 0.4325
- (b) 0.9529
- (c) 0.5675
- (d) 0.04758. Effect of Sample Size on Analysis
- If the random sample size were n = 5:
- Can you answer question 9?
- (a) Yes
- (b) No9. Normal Distribution Assumption
- If sample size were n = 5 and the distribution of lifetimes follows a Normal model:
- Can we answer question 9?
- (a) Yes
- (b) No
Problem Set 4: Confidence Intervals
10. Confidence Interval for NIU Student Opinions
- In a sample of 225 NIU students, 99 believe professors shouldn't give finals early.
- Calculate the 95% confidence interval for the proportion:
- (a) 0.44 ± 0.033
- (b) 0.44 ± 0.075
- (c) 0.44 ± 0.065
- (d) 0.44 ± 0.01511. Job Finding Statistics for NIU Graduates
- Sample size of 625 graduates, confidence interval: 0.68 to 0.72.
- Determine which statement is true:
- (a) 95% chance sample proportion is between 0.68 and 0.72.
- (b) 95% chance population proportion is 0.70.
- (c) 95% confidence sample proportion is between 0.68 and 0.72.
- (d) 95% confidence population proportion is between 0.68 and 0.72.12. Confidence Interval for High Achieving Students Teaching
- Gallup poll: 76% of Americans support recruiting high-achieving students for teaching.
- Determine the 95% confidence interval:
- (a) 0.70 to 0.82
- (b) 0.73 to 0.79
- (c) 0.33 to 1.19
- (d) 0.76 to 0.9513. Interpretation of Confidence Intervals
- Analyze the correct confidence interval's conclusions:
- (a) 95% confidence true proportion is 0.76.
- (b) 95% sure sample proportion from 200 Americans is 0.76.
- (c) 95% confidence true proportion is within interval.
- (d) We know for sure true proportion within interval.
Problem Set 5: Hypothesis Testing
14. Null and Alternative Hypotheses for Student Study
- 1996 study: 31% of students reported mothers graduated from college.
- Recent study shows 33%. Determine null and alternative hypotheses:
- (a) $H_0: p = 0.31$ vs $H_A: p
eq 0.31$
- (b) $H_0: p = 0.33$ vs $H_A: p
eq 0.33$
- (c) $H_0: p = 0.31$ vs $H_A: p > 0.31$
- (d) $H_0: p = 0.33$ vs $H_A: p < 0.33$15. Test Statistic for Seed Germination Rate
- Packet claims 92% seeds germinate. Observed: 180 out of 200 seeds germinate.
- Find the appropriate test statistic, z:
- (a) −3.39
- (b) 0.90
- (c) −0.092
- (d) −1.0416. Finding p-value from Test Statistic
- If testing Alternative Hypothesis that a proportion is greater than expected and observed test statistic = 0.96, find p-value:
- (a) 0.3315
- (b) 0.1685
- (c) 0.3370
- (d) 0.674017. Decision from Hypothesis Test with p-value
- p-value 0.0213 leads to appropriate decision:
- (a) Reject the Null Hypothesis; evidence supports Alternative Hypothesis.
- (b) Reject the Alternative Hypothesis; evidence supports Null Hypothesis.
- (c) Do not reject Null Hypothesis; no evidence for Alternative Hypothesis.
- (d) Do not reject Alternative Hypothesis; no evidence for Null Hypothesis.18. Interpretation of Germination Rate Decision
- For seeds, if gardener believes germination rate is less than 92% and fails to reject Null Hypothesis:
- (a) Strong evidence data shows germination rate less than 92%.
- (b) We know disease rate is less than 92%.
- (c) Germination rate is shown to be equal to 92%.
- (d) Not enough evidence to conclude germination rate is less than 92%.