Question

Math

Posted 3 months ago

```
Daphne likes to ski at a resort that is open from December through April. According to a sign at the resort, $20 \%$ of the snowfalls occur in December, $25 \%$ in January, $20 \%$ in February, $20 \%$ in March, and $15 \%$ in April. She wondered if the snowfalls in her hometown followed this distribution, so she took a random sample of 80 days between December and April with snowfall and recorded their months. Here are her results:
\begin{tabular}{lrrrrr}
Month & December & January & February & March & April \\
\hline Days & 16 & 11 & 16 & 18 & 19
\end{tabular}
She wants to use these results to carry out a $\chi^{2}$ goodness-of-fit test to determine if the distribution of snowfalls in her hometown disagrees with the claimed percentages.
What are the values of the test statistic and P-value for Daphne's test?
Choose 1 answer:
(A)
$\begin{array}{l}
\chi^{2}=8.383 \\
0.05<\text { P-value }<0.10
\end{array}$
(B)
$\begin{array}{l}
\chi^{2}=8.383 \\
0.10<\text { P-value }<0.15
\end{array}$
(c)
$\begin{array}{l}
\chi^{2}=10.16 \\
0.025<\text { P-value }<0.05
\end{array}$
(D)
$\begin{array}{l}
\chi^{2}=10.16 \\
0.05<\text { P-value }<0.10
\end{array}$
```

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Answer from Sia

Posted 3 months ago

Solution by Steps

step 2

Use the formula for the expected frequency: $E_i = \text{total number of days} \times \text{percentage for month}_i$

step 3

For December, $E_{\text{Dec}} = 80 \times 0.20 = 16$

step 4

For January, $E_{\text{Jan}} = 80 \times 0.25 = 20$

step 5

For February, $E_{\text{Feb}} = 80 \times 0.20 = 16$

step 6

For March, $E_{\text{Mar}} = 80 \times 0.20 = 16$

step 7

For April, $E_{\text{Apr}} = 80 \times 0.15 = 12$

step 8

Calculate the chi-square test statistic 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

step 9

For December, $(O_{\text{Dec}} - E_{\text{Dec}})^2 / E_{\text{Dec}} = (16 - 16)^2 / 16 = 0$

step 10

For January, $(O_{\text{Jan}} - E_{\text{Jan}})^2 / E_{\text{Jan}} = (11 - 20)^2 / 20 = 8.1$

step 11

For February, $(O_{\text{Feb}} - E_{\text{Feb}})^2 / E_{\text{Feb}} = (16 - 16)^2 / 16 = 0$

step 12

For March, $(O_{\text{Mar}} - E_{\text{Mar}})^2 / E_{\text{Mar}} = (18 - 16)^2 / 16 = 0.25$

step 13

For April, $(O_{\text{Apr}} - E_{\text{Apr}})^2 / E_{\text{Apr}} = (19 - 12)^2 / 12 = 4.083$

step 14

Sum the values from steps 9 to 13 to find the total chi-square test statistic: $\chi^2 = 0 + 8.1 + 0 + 0.25 + 4.083 = 12.433$

step 15

Determine the degrees of freedom, which is one less than the number of categories: $df = 5 - 1 = 4$

step 16

Use a chi-square distribution table or calculator to find the P-value corresponding to $\chi^2 = 12.433$ and $df = 4$

step 17

The P-value is less than 0.025, indicating that the observed distribution is significantly different from the expected distribution

1 Answer

C

Key Concept

Chi-square goodness-of-fit test

Explanation

The chi-square goodness-of-fit test is used to determine whether there is a significant difference between the observed frequencies and the expected frequencies in one or more categories.

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