Compute the expected counts for each color.
Color
Frequency
Expected Count
Brown
62

51.48
Yellow
64

55.44
Red
57

51.48
Blue
61

95.04
Orange
89

79.20
Green
63

63.36
​(Round to two decimal places as​ needed.)
Part 3
What is the test​ statistic?
chi Subscript 0 Superscript 2
equals

enter your response here
​(Round to three decimal places as​ needed.)

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?**

---

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).

Learn how to compute expected counts for different colors and calculate the chi-squared test statistic for a given dataset. This problem involves applying the chi-squared test for goodness of fit, using observed and expected frequencies to analyze data.

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