Guided Practice: Comparing Means and Confidence Intervals in Statistics

Q1. Construct and interpret a 98% confidence interval for the difference in mean returns between a growth portfolio and an income portfolio.

Background

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

This question tests your ability to construct and interpret a confidence interval for the difference in means from two independent samples. This is a common inferential statistics technique used to estimate how much two population means differ, based on sample data.

Key Terms and Formulas

• Confidence Interval (CI): A range of values, derived from sample statistics, that is likely to contain the true population parameter.

• Difference in Means: $\mu_1 - \mu_2$ (here, growth portfolio mean minus income portfolio mean)

• Standard Error (SE) for Difference in Means:

$SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$

• Critical Value ($z^*$ or $t^*$): For large samples, you can use the $z$-distribution. For a 98% CI, find the appropriate $z^*$ value.

• Confidence Interval Formula:

$\left[ (\bar{x}_1 - \bar{x}_2) \pm z^* \times SE \right]$

Step-by-Step Guidance

1. Identify the sample statistics for both groups: Growth Portfolio: $n_1 = 50$, $\bar{x}_1 = 8\%$, $s_1 = 10\%$ Income Portfolio: $n_2 = 60$, $\bar{x}_2 = 6\%$, $s_2 = 12\%$

2. Calculate the point estimate for the difference in means: $\bar{x}_1 - \bar{x}_2 = 8\% - 6\%$

3. Compute the standard error (SE) for the difference in means: $SE = \sqrt{\frac{(10)^2}{50} + \frac{(12)^2}{60}}$

4. Find the critical value $z^*$ for a 98% confidence interval. (Hint: For 98%, $z^*$ is typically found using a standard normal table.)

5. Set up the confidence interval formula: $\left[ (\bar{x}_1 - \bar{x}_2) \pm z^* \times SE \right]$

Try solving on your own before revealing the answer!

Final Answer:

The 98% confidence interval for the difference in mean returns is approximately [−1.13%, 5.13%].

This means we are 98% confident that the true difference in average annual returns (growth minus income) falls within this interval.

Q2. Test if there is a significant difference in mean productivity between Silicon Valley and Austin offices at a 5% significance level.

Background

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

This question is about conducting a two-sample hypothesis test to determine if there is a statistically significant difference between the means of two independent groups.

Key Terms and Formulas

• Null Hypothesis ($H_0$): $\mu_1 = \mu_2$ (no difference in means)

• Alternative Hypothesis ($H_a$): $\mu_1 \neq \mu_2$ (means are different)

• Test Statistic (z or t):

$z = \frac{(\bar{x}_1 - \bar{x}_2)}{SE}$

• Standard Error (SE):

$SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$

• Significance Level ($\alpha$): 0.05 (5%)

• Critical Value: For a two-tailed test at $\alpha = 0.05$, the critical $z$-values are $\pm 1.96$.

Step-by-Step Guidance

1. State the hypotheses: $H_0: \mu_1 = \mu_2$ $H_a: \mu_1 \neq \mu_2$

2. List the sample statistics: Silicon Valley: $n_1 = 40$, $\bar{x}_1 = 150$, $s_1 = 20$ Austin: $n_2 = 40$, $\bar{x}_2 = 140$, $s_2 = 25$

3. Calculate the standard error (SE): $SE = \sqrt{\frac{20^2}{40} + \frac{25^2}{40}}$

4. Compute the test statistic: $z = \frac{150 - 140}{SE}$

5. Compare the calculated $z$ value to the critical values ($\pm 1.96$) to determine if you reject or fail to reject $H_0$.

Try solving on your own before revealing the answer!

Final Answer:

The calculated test statistic is approximately 1.74, which does not exceed the critical value of 1.96. Therefore, there is not enough evidence to conclude a significant difference in mean productivity at the 5% level.