Confidence Intervals Summary

Confidence Intervals

Overview

• Use sample mean $M$ to estimate population mean $\mu$.
• Assess the accuracy of $M$ as an estimator of $\mu$.

Sampling Distribution of Mean

• The mean of the sampling distribution is $\mu$.
• The standard deviation is $\sigma_M = \frac{\sigma}{\sqrt{n}}$, also known as the standard error of the mean.
• Shape: normal distribution.

Using Sampling Distribution

• For a population with $\mu=100$ and $\sigma=10$, with a sample size of $n=25$, the sampling distribution of the mean is approximately normal.
• Mean = 100
• Standard deviation = $\sigma_M = \frac{10}{\sqrt{25}} = 2$

Area Under the Curve

• The sampling distribution is an idealized frequency distribution.
• Area under the curve corresponds to the proportion of cases in a particular range of scores and the probability of sampling a case within that range.

Normal Distribution

• Symmetrical.
• The mean is the center.
• 50% of the distribution is above and below the mean.
• The area between any two points can be calculated.

Z-Scores

• Convert raw scores to Z scores to use the normal curve table.
• $Z = \frac{M - \mu}{\sigma_M}$

Examples of Probability Calculations:

• Example 1: Probability of a sample mean higher than 103
  • Calculate $\sigma_M = 2$
  • Calculate Z score: $Z = \frac{103 - 100}{2} = 1.5$
  • From tables, the area beyond $z = 1.5$ is $0.5 - 0.433 = 0.067$.
  • Therefore, the probability is 0.067.
• Example 2: Probability that a sample mean will be within $\pm 4$ points of $\mu$
  • Calculate $\sigma_M = 2$
  • Convert 96 and 104 to Z scores: $Z = \frac{96 - 100}{2} = -2$, $Z = \frac{104 - 100}{2} = 2$
  • From tables, the area between $Z = \pm 2$ is $0.477 \times 2 = 0.954$.
  • Therefore, the probability is 0.954.
• Example 3: Value exceeded by only 2.5% of sample means
  • Calculate $\sigma_M = 2$
  • From tables, the Z score that corresponds to an area of 0.475 is 1.96.
  • Convert Z to M: $M = \mu + Z \sigma_M = 100 + 1.96 \times 2 = 103.92$
• Example 4: Limits within which the central 95% of sample means fall
  • Calculate $\sigma_M = 2$
  • From tables, $z = \pm 1.96$ bounds the central area of 95%.
  • Upper limit: $M<em>{upper} = \mu + z \sigma</em>M = 100 + 1.96 \times 2 = 103.92$
  • Lower limit: $M<em>{lower} = \mu - z \sigma</em>M = 100 - 1.96 \times 2 = 96.08$

Estimation Concepts

• Point Estimate:
  • The sample mean $M$ is the best point estimate of $\mu$, as it is an unbiased estimator.
• Interval Estimate (Confidence Interval):
  • Construct a range of values that, with a certain level of confidence, covers the true value of $\mu$.
  • If 95% of sample means are within $1.96 \times \sigma<em>M$ from $\mu$, then for 95% of samples, $\mu$ is within $1.96 \times \sigma</em>M$ from $M$.

Confidence Interval for $\mu$ ($\sigma$ known)

• Let $\alpha$ = error rate as a probability
  • For 95% confidence, $\alpha = 0.05$
  • For 90% confidence, $\alpha = 0.10$
  • For 99% confidence, $\alpha = 0.01$
• A $100(1 - \alpha)\%$ confidence interval for $\mu$ has limits:
  • $\mu<em>{upper} = M + (z</em>c \times \sigma_M)$
  • $\mu<em>{lower} = M - (z</em>c \times \sigma_M)$
  • Where $z_c$ = “critical value”: the value of z that cuts off $\alpha/2$ in each tail of the distribution, e.g., 1.96 for $\alpha = 0.05$.

Finding $z_c$

• For a 90% confidence interval, find the Z score that corresponds to an area of 0.45 (since 5% is in each tail).

Example

• Scores on a statistics exam are normally distributed with $\sigma^2 = 144$. Let $n = 25$ and $M = 60$. Obtain 90% confidence limits for $\mu$.
  • Step 1: Find $z<em>c$ and $\sigma</em>M$
    • $z_c = 1.645$
    • $\sigma = \sqrt{144} = 12$
    • $\sigma_M = \frac{12}{\sqrt{25}} = 2.4$
  • Step 2: Find upper and lower limits for $\mu$
    • $\mu<em>{upper} = M + (z</em>c \times \sigma_M) = 60 + (1.645 \times 2.4) = 63.95$
    • $\mu<em>{lower} = M - (z</em>c \times \sigma_M) = 60 - (1.645 \times 2.4) = 56.05$
  • Step 3: Conclusion
    • We can be 90% confident that the population mean exam mark lies between 56.05 and 63.95.

Effect of Confidence Level on Width of Interval

• Calculate 99% confidence limits for $\mu$ for statistics exam data. $n = 25$, $M = 60$, $\sigma_M = 2.4$.
  • From tables, $Z_C = 2.58$ (area = 0.495).
  • Upper and lower limits for $\mu$
    • $\mu<em>{upper} = M + (z</em>c \times \sigma_M) = 60 + (2.58 \times 2.4) = 66.19$
    • $\mu<em>{lower} = M - (z</em>c \times \sigma_M) = 60 - (2.58 \times 2.4) = 53.81$
  • Compare these with 90% confidence limits [56.05, 63.95]. The greater the level of confidence desired, the wider the interval.