A gym owner wants to know if the gym has similar numbers of members across different age groups. The table shows the distribution of ages for members from a random survey. Write the null & alt. hypotheses to test the claim that the gym has equal numbers of members across all age groups.

• A. H₀: The # of members is the same for all age groupsH_a: The # of members is significantly different between the age groups
• B. H₀: The # of members is the same for all age groupsH_a: The # of members is significantly different for at least one of the age groups
• C. H₀: The # of members is significantly different for at least one of the age groupsH_a: The # of members is the same for all age groups
• D. H₀: The # of members is significantly different between the age groupsH_a: The # of members is the same for all age groups

Answer: (B)

Verified step by step guidance

1. Step 1: Understand the problem. The gym owner wants to test whether the number of members is evenly distributed across different age groups. This is a hypothesis testing problem where we compare observed frequencies to expected frequencies.
2. Step 2: Define the null and alternative hypotheses. The null hypothesis (H0) states that the number of members is the same across all age groups. The alternative hypothesis (Ha) states that the number of members is significantly different for at least one of the age groups.
3. Step 3: Choose the appropriate statistical test. Since we are comparing observed frequencies across categories, a Chi-Square Goodness-of-Fit test is appropriate for this problem.
4. Step 4: Calculate the expected frequencies. If the null hypothesis is true, the gym members should be evenly distributed across all age groups. Calculate the expected frequency for each age group by dividing the total number of members by the number of age groups.
5. Step 5: Compute the Chi-Square test statistic. Use the formula: χ² = Σ((Oᵢ - Eᵢ)² / Eᵢ), where Oᵢ is the observed frequency and Eᵢ is the expected frequency for each age group. Then compare the test statistic to the critical value from the Chi-Square distribution table at the chosen significance level to determine whether to reject or fail to reject the null hypothesis.