Business Statistics: Hypothesis Testing, Goodness-of-Fit, and Contingency Table Guidance

Q2. Do the observed cell phone number frequencies contradict the guide's reported percentages?

Background

Topic: Goodness-of-Fit Test (Multinomial Experiments)

This question tests your ability to use the chi-square goodness-of-fit test to compare observed frequencies with expected frequencies based on hypothesized proportions.

Key Terms and Formulas

• Observed Frequency ($n_i$): The actual count in each category from the sample.

• Expected Frequency ($E_i$): The count expected in each category if the null hypothesis is true.

• Hypothesized Proportion ($p_{i0}$): The proportion claimed by the guide for each category.

• Test Statistic: $\chi^2 = \sum_{i=1}^k \frac{(n_i - E_i)^2}{E_i}$

• Expected Frequency: $E_i = n p_{i0}$

• Degrees of Freedom: $df = k - 1$

Step-by-Step Guidance

1. State the null and alternative hypotheses:

   • Null hypothesis ($H_0$): The observed frequencies match the guide's reported proportions.

   • Alternative hypothesis ($H_1$): The observed frequencies do not match the guide's reported proportions.

2. List the hypothesized proportions and observed frequencies for each category:

   • One number: $p_1 = 0.52$, observed = 472

   • Two numbers: $p_2 = 0.34$, observed = 331

   • Three numbers: $p_3 = 0.08$, observed = 111

   • Four or more: $p_4 = 0.06$, observed = 34

3. Calculate the expected frequency for each category using $E_i = n p_{i0}$, where $n = 948$.

   • For example, for one number: $E_1 = 948 \times 0.52$

   • Repeat for each category.

4. Compute the chi-square test statistic using:

   $\chi^2 = \sum_{i=1}^4 \frac{(n_i - E_i)^2}{E_i}$

   Calculate the squared difference for each category, divide by the expected frequency, and sum across all categories.

5. Determine the degrees of freedom: $df = k - 1$, where $k$ is the number of categories.

Try solving on your own before revealing the answer!

Q3. Do fatal DWI crashes occur equally across all days of the week?

Background

Topic: Goodness-of-Fit Test (Uniform Distribution)

This question asks you to test whether the number of fatal DWI crashes is distributed equally across the seven days of the week using the chi-square goodness-of-fit test.

Key Terms and Formulas

• Observed Frequency ($n_i$): Number of crashes for each day.

• Expected Frequency ($E_i$): If crashes are equally likely, $E_i = \frac{n}{k}$ for each day.

• Test Statistic: $\chi^2 = \sum_{i=1}^k \frac{(n_i - E_i)^2}{E_i}$

• Degrees of Freedom: $df = k - 1$

Step-by-Step Guidance

1. State the null and alternative hypotheses:

   • Null hypothesis ($H_0$): Crashes are equally likely on each day.

   • Alternative hypothesis ($H_1$): Crashes are not equally likely on each day.

2. List the observed frequencies for each day (Sunday through Saturday).

3. Calculate the expected frequency for each day: $E_i = \frac{217}{7}$

4. Compute the chi-square test statistic:

   $\chi^2 = \sum_{i=1}^7 \frac{(n_i - E_i)^2}{E_i}$

5. Determine the degrees of freedom: $df = 7 - 1$

Try solving on your own before revealing the answer!

Q4. Is brand loyalty independent of the origin of the car manufacturer?

Background

Topic: Test of Independence (Contingency Table Analysis)

This question tests your ability to use the chi-square test for independence to determine if two categorical variables (brand loyalty and manufacturer origin) are independent.

Key Terms and Formulas

• Observed Frequency ($n_{ij}$): The count in each cell of the contingency table.

• Expected Frequency ($E_{ij}$): $E_{ij} = \frac{R_i C_j}{n}$, where $R_i$ is the row total, $C_j$ is the column total, and $n$ is the grand total.

• Test Statistic: $\chi^2 = \sum_{i} \sum_{j} \frac{(n_{ij} - E_{ij})^2}{E_{ij}}$

• Degrees of Freedom: $df = (r-1)(c-1)$, where $r$ is the number of rows and $c$ is the number of columns.

Step-by-Step Guidance

1. State the null and alternative hypotheses:

   • Null hypothesis ($H_0$): Brand loyalty is independent of manufacturer origin.

   • Alternative hypothesis ($H_1$): Brand loyalty is not independent of manufacturer origin.

2. Set up the observed frequency table with the given data for each combination of brand loyalty and manufacturer origin.

3. Calculate the row totals, column totals, and grand total ($n$).

4. Compute the expected frequency for each cell using $E_{ij} = \frac{R_i C_j}{n}$.

5. Calculate the chi-square test statistic:

   $\chi^2 = \sum_{i} \sum_{j} \frac{(n_{ij} - E_{ij})^2}{E_{ij}}$

6. Determine the degrees of freedom: $df = (r-1)(c-1)$

Try solving on your own before revealing the answer!