To compute the test statistic for the given data, we will use the chi-squared test for goodness of fit. The chi-squared test statistic is calculated using the formula:

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

where $O_i$ is the observed frequency and $E_i$ is the expected frequency.

Given dataTo compute the test statistic for the given data, we will use the chi-squared test for goodness of fit. The chi-squared test statistic is calculated using the formula:

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

where $O_i$ is the observed frequency and $E_i$ is the expected frequency.

Given

• Brown: $O_1 = 62$, $E_1 = 51.48$
• Yellow: $O_2 = 64$, $E_2 = 55.44$
• Red: $O_3 = 57$, $E_3 = 51.48$
• Blue: $O_4 = 61$, $E_4 = 95.04$
• Orange: $O_5 = 89$, $E_5 = 79.20$
• Green: $O_6 = 63$, $E_6 = 63.36$

Now, calculate the chi-squared statistic:

$\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} = \frac{(62 - 51.48)^2}{51.48} + \frac{(64 - 55.44)^2}{55.44} + \frac{(57 - 51.48)^2}{51.48} + \frac{(61 - 95.04)^2}{95.04} + \frac{(89 - 79.20)^2}{79.20} + \frac{(63 - 63.36)^2}{63.36}$

Let's compute each term separately:

$\frac{(62 - 51.48)^2}{51.48} = \frac{(10.52)^2}{51.48} = \frac{110.6704}{51.48} \approx 2.150$

$\frac{(64 - 55.44)^2}{55.44} = \frac{(8.56)^2}{55.44} = \frac{73.2736}{55.44} \approx 1.322$

$\frac{(57 - 51.48)^2}{51.48} = \frac{(5.52)^2}{51.48} = \frac{30.4704}{51.48} \approx 0.592$

$\frac{(61 - 95.04)^2}{95.04} = \frac{(-34.04)^2}{95.04} = \frac{1158.7216}{95.04} \approx 12.192$

$\frac{(89 - 79.20)^2}{79.20} = \frac{(9.8)^2}{79.20} = \frac{96.04}{79.20} \approx 1.213$

$\frac{(63 - 63.36)^2}{63.36} = \frac{(-0.36)^2}{63.36} = \frac{0.1296}{63.36} \approx 0.002$

Summing these values:

$\chi^2 \approx 2.150 + 1.322 + 0.592 + 12.192 + 1.213 + 0.002 \approx 17.471$

So, the test statistic $\chi^2$ is approximately:

$\chi^2 = 17.471$

Would you like more details or have any questions?

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Here are 8 relative questions that expand on this information:

1. What is the chi-squared test used for in statistical analysis?
2. How do you determine the degrees of freedom for a chi-squared test?
3. What is the critical value for a chi-squared test with 5 degrees of freedom at a significance level of 0.05?
4. Can the chi-squared test be used for data that are not frequencies or counts?
5. What are the assumptions required for a chi-squared test to be valid?
6. How do you interpret the p-value obtained from a chi-squared test?
7. What are some common applications of the chi-squared test in real-world scenarios?
8. How would you handle expected counts that are very low in a chi-squared test?

Tip: Always check the assumptions of the chi-squared test before applying it, such as ensuring that the expected frequencies are sufficiently large (typically at least 5).