A doctor keeps track of the number of babies she delivers in each season. She expects that the distribution will be uniform (the same number of babies in each season). The data she collects is shown in the table below. Find the test statistic, [tex]\chi_0^2[/tex], for the chi-square goodness-of-fit test. Round the final answer to one decimal place.

[tex]\chi_0^2=\sum \frac{(O-E)^2}{E}[/tex]

\begin{tabular}{|c|c|c|c|c|}
\hline Season & Spring & Summer & Fall & Winter \\
\hline Expected & 28 & 28 & 28 & 28 \\
\hline Observed & 35 & 21 & 28 & 28 \\
\hline
\end{tabular}

Provide your answer below:
chi-square test statistic [tex]= \square[/tex]



Answer :

To determine the chi-square test statistic, [tex]\(\chi_0^2\)[/tex], we follow the formula:
[tex]\[ \chi_0^2 = \sum \frac{(O - E)^2}{E} \][/tex]
where [tex]\(O\)[/tex] represents the observed frequencies and [tex]\(E\)[/tex] represents the expected frequencies. In this case, we have:

- Observed frequencies: [tex]\(O = [35, 21, 28, 28]\)[/tex]
- Expected frequencies: [tex]\(E = [28, 28, 28, 28]\)[/tex]

Let's calculate the chi-square test statistic step by step:

1. For Spring:
[tex]\[ \frac{(35 - 28)^2}{28} = \frac{7^2}{28} = \frac{49}{28} = 1.75 \][/tex]

2. For Summer:
[tex]\[ \frac{(21 - 28)^2}{28} = \frac{(-7)^2}{28} = \frac{49}{28} = 1.75 \][/tex]

3. For Fall:
[tex]\[ \frac{(28 - 28)^2}{28} = \frac{0^2}{28} = 0 \][/tex]

4. For Winter:
[tex]\[ \frac{(28 - 28)^2}{28} = \frac{0^2}{28} = 0 \][/tex]

Now, summing these values:
[tex]\[ 1.75 + 1.75 + 0 + 0 = 3.5 \][/tex]

Thus, the chi-square test statistic is:
[tex]\[ \chi_0^2 = 3.5 \][/tex]