Let's go step by step through the problem in the image.

This problem is related to a Chi-square ($\chi^2$) goodness-of-fit test. The data presented has observed frequencies and expected frequencies for different categories (A, B, C, D, E).

Step 1: Calculate the $\chi^2$ test-statistic.

The formula for the $\chi^2$ test-statistic is:

$\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$

Where:

• $O_i$ = observed frequency in each category
• $E_i$ = expected frequency in each category

Given dataLet's go step by step through the problem in the image.

This problem is related to a Chi-square ($\chi^2$) goodness-of-fit test. The data presented has observed frequencies and expected frequencies for different categories (A, B, C, D, E).

Step 1: Calculate the $\chi^2$ test-statistic.

The formula for the $\chi^2$ test-statistic is:

$\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$

Where:

• $O_i$ = observed frequency in each category
• $E_i$ = expected frequency in each category

Given

Let's calculate the $\chi^2$ value:

For D: $\frac{(5 - 10.8)^2}{10.8} = \frac{(5.8)^2}{10.8} = \frac{33.64}{10.8} \approx 3.1148$

For E: $\frac{(21 - 10.8)^2}{10.8} = \frac{(10.2)^2}{10.8} = \frac{104.04}{10.8} \approx 9.63$

Now summing all the values together:

$\chi^2 = 1.337 + 0.1333 + 0.3 + 3.1148 + 9.63 = 14.5151$

Step 2: Finding the p-value.

To find the p-value, we need the degrees of freedom (df):

$\text{df} = \text{number of categories} - 1 = 5 - 1 = 4$

Using a Chi-square distribution table or calculator, with $\chi^2 = 14.5151$ and $df = 4$, the corresponding p-value is approximately 0.0057.

Step 3: Conclusion of the hypothesis test.

The hypothesis for a chi-square test is typically:

• Null hypothesis ($H_0$): The observed frequencies match the expected frequencies.
• Alternative hypothesis ($H_A$): The observed frequencies do not match the expected frequencies.

Since the p-value (0.0057) is less than the common significance level (0.05), we reject the null hypothesis.

Step 4: Final conclusion.

The rejection of the null hypothesis means there is sufficient evidence to conclude that the test question responses do not occur with the same frequency.

Would you like more details on any part of the explanation?

Here are 5 related questions you could consider:

1. How do we determine the degrees of freedom for a chi-square test?
2. How can we find the p-value using a chi-square distribution table?
3. What are some assumptions behind the chi-square goodness-of-fit test?
4. How does the significance level affect the conclusion of hypothesis tests?
5. What is the relationship between observed and expected frequencies in this test?

Tip: When using a Chi-square test, always ensure the expected frequencies are large enough (typically at least 5) to avoid inaccuracies.