**Question**

256 musicians were polled to determine their zodaic sign. The results are given below

Aries 29

Torus 24

Gemini 22

Cancer 19

Leo 21

Virgo 18

Libra 19

Scorpio 20

Sagittarius 23

Capricorn 18

Aquarius 20

Pisces 23

Test the hypothesis that the zodaic signs are evenly distributed accross the musicians.

**Solution Steps**

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The formular for chi-square is

O = Observed values

E = Expected value

**Step 1: State the null and alternate hypothesis**

**H**The values are evenly distributed

_{0}:**H**: The values are not evenly distributed

_{a}Acceptance criteria for the null value is for large p-Value. (we would calculate this)

Rejection criteria is when p-value is low.

**Step 2:Create the Table**

I have created the table in Excel as shown below

**Step 3: Calculate the Expected Value**

The expected value corresponds to the mean of the samples and has been calculated as 21.33.

**Step 4: Calculate the Difference**

This is given by the formular Observed Value - Expected Value (O - E) for each of the observations.

**Step 5: Calculate the Squared Difference**

This is the square of the value calculated in Step 4. The values are shown in Table 3.

**Step 6: Calculate the Component**

This is given by the chi-square formula

At this point the excel sheet would look like the one below

**Table 3**: Final Chi-Square Table

**Step 7: Calculate the Chi-Square Statitic**

You can easily find this value by taking the sum of the last column.

Chi-Square test statistic = 5.09

**Step 8: Find the p-Value**

You can find the p-value using the calculate chi-square statistic and the degrees of freedom

Degrees of freedom is given by N-1

df = 12 - 1 = 11

where N is the number of observations in this case

See statistical tables here

From the chi-square table we get a p-value of between

**0.900**and

**0.950.**

**Step 9: State the Conclustion**

In chi-square, the null-hypotheis is accepted if the p-value is very large, say 90% to 100% and we reject the null hypothesis for small values of p-value.

In this case, the p-value is between 90% and 95%. Therefore, we accept the null hypothesis.

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