Practice Exam Guidance: Sampling Distributions, Confidence Intervals, and Normal Probability

Q1. When is the sampling distribution of all sample means normally distributed?

Background

Topic: Central Limit Theorem & Sampling Distributions

This question tests your understanding of the conditions under which the sampling distribution of sample means is normal. The Central Limit Theorem (CLT) is a key concept here.

Key Terms:

• Central Limit Theorem (CLT): States that the sampling distribution of the sample mean approaches a normal distribution as the sample size increases, regardless of the population's distribution, provided certain conditions are met.

• Sample Size: The number of observations in a sample.

• Population Distribution: The distribution from which the sample is drawn.

Step-by-Step Guidance

1. Recall the Central Limit Theorem: For large sample sizes (usually $n \geq 30$), the sampling distribution of the sample mean is approximately normal, even if the population distribution is not normal.

2. If the population distribution is already normal, then the sampling distribution of the sample mean is normal for any sample size.

3. Consider the answer choices: Which options reflect these two conditions?

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Q2. If the sample size is increased, how is margin of error for a confidence interval affected, everything else being equal?

Background

Topic: Margin of Error & Confidence Intervals

This question tests your understanding of how sample size impacts the margin of error in confidence intervals.

Key Terms and Formula:

• Margin of Error ($E$): The range within which the true population parameter is expected to lie.

• Sample Size ($n$): Number of observations in the sample.

• Standard Error ($SE$): $SE = \frac{\sigma}{\sqrt{n}}$

• Margin of Error Formula: $E = z^* \cdot SE$

Step-by-Step Guidance

1. Recall that the margin of error depends on the standard error, which is inversely related to the square root of the sample size.

2. As sample size increases, $\sqrt{n}$ increases, so $SE$ decreases.

3. Therefore, the margin of error $E$ decreases as sample size increases, all else being equal.

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Q3. The graph of a normal curve is given. If one individual value is selected at random, what is the probability that it is greater than 849?

Background

Topic: Normal Distribution & Probability

This question tests your ability to interpret probabilities from a normal distribution graph. The value 849 is the mean, and the normal curve is symmetric about the mean.

Key Terms:

• Normal Distribution: A symmetric, bell-shaped distribution characterized by its mean and standard deviation.

• Mean: The center of the distribution (here, 849).

• Probability: The area under the curve to the right of 849 represents the probability of selecting a value greater than 849.

Step-by-Step Guidance

1. Identify that 849 is the mean of the normal distribution.

2. Recall that the normal distribution is symmetric about the mean.

3. The probability of selecting a value greater than the mean is equal to the area under the curve to the right of the mean.

4. For a normal distribution, this area is exactly half of the total area under the curve.

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Q4. A random sample of 1493 adults was asked “Are you afraid to go out at night within a mile of your home because of crime?” Of that sample, 672 say “Yes.”

Background

Topic: Estimating Proportions & Confidence Intervals

This question tests your ability to estimate population proportions, check normality conditions, and construct confidence intervals.

Key Terms and Formulas:

• Point Estimate ($\hat{p}$): $\hat{p} = \frac{x}{n}$, where $x$ is the number of successes and $n$ is the sample size.

• Sampling Distribution: The distribution of sample proportions.

• Normality Condition: $n\hat{p} \geq 5$ and $n(1-\hat{p}) \geq 5$

• Confidence Interval for Proportion: $\hat{p} \pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

Step-by-Step Guidance

1. Calculate the point estimate: $\hat{p} = \frac{672}{1493}$

2. Check normality: Compute $n\hat{p}$ and $n(1-\hat{p})$ to ensure both are at least 5.

3. Find the critical value $z^*$ for an 86% confidence interval (look up or calculate).

4. Set up the confidence interval formula: $\hat{p} \pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

5. Interpret the interval: Explain what the confidence interval means in context.

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Q5. A newspaper reported that 21% of all adults in the U.S. are illiterate. Describe the sampling distribution of sample proportions for samples of 125 adults.

Background

Topic: Sampling Distribution of Proportions

This question tests your ability to describe the mean, standard deviation, and shape of the sampling distribution for sample proportions.

Key Terms and Formulas:

• Population Proportion ($p$): 0.21

• Sample Size ($n$): 125

• Mean of Sampling Distribution: $\mu_{\hat{p}} = p$

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

• Normality Condition: $np \geq 5$ and $n(1-p) \geq 5$

Step-by-Step Guidance

1. Calculate the mean: $\mu_{\hat{p}} = 0.21$

2. Calculate the standard deviation: $\sigma_{\hat{p}} = \sqrt{\frac{0.21 \times 0.79}{125}}$

3. Check normality: $125 \times 0.21$ and $125 \times 0.79$ should both be at least 5.

4. Describe the shape: If normality conditions are met, the distribution is approximately normal.

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