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Math
Posted 10 months ago

Daphne likes to ski at a resort that is open from December through April. According to a sign at the resort, 20%20 \% of the snowfalls occur in December, 25%25 \% in January, 20%20 \% in February, 20%20 \% in March, and 15%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 χ2\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)
χ2=8.3830.05< P-value <0.10
\begin{array}{l}
\chi^{2}=8.383 \\
0.05<\text { P-value }<0.10
\end{array}

(B)
χ2=8.3830.10< P-value <0.15
\begin{array}{l}
\chi^{2}=8.383 \\
0.10<\text { P-value }<0.15
\end{array}

(c)
χ2=10.160.025< P-value <0.05
\begin{array}{l}
\chi^{2}=10.16 \\
0.025<\text { P-value }<0.05
\end{array}

(D)
χ2=10.160.05< P-value <0.10
\begin{array}{l}
\chi^{2}=10.16 \\
0.05<\text { P-value }<0.10
\end{array}
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Answer from Sia
Posted 10 months ago
Solution by Steps
step 2
Use the formula for the expected frequency: Ei=total number of days×percentage for monthiE_i = \text{total number of days} \times \text{percentage for month}_i
step 3
For December, EDec=80×0.20=16E_{\text{Dec}} = 80 \times 0.20 = 16
step 4
For January, EJan=80×0.25=20E_{\text{Jan}} = 80 \times 0.25 = 20
step 5
For February, EFeb=80×0.20=16E_{\text{Feb}} = 80 \times 0.20 = 16
step 6
For March, EMar=80×0.20=16E_{\text{Mar}} = 80 \times 0.20 = 16
step 7
For April, EApr=80×0.15=12E_{\text{Apr}} = 80 \times 0.15 = 12
step 8
Calculate the chi-square test statistic using the formula: χ2=∑(Oi−Ei)2Ei\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}, where OiO_i is the observed frequency and EiE_i is the expected frequency
step 9
For December, (ODec−EDec)2/EDec=(16−16)2/16=0(O_{\text{Dec}} - E_{\text{Dec}})^2 / E_{\text{Dec}} = (16 - 16)^2 / 16 = 0
step 10
For January, (OJan−EJan)2/EJan=(11−20)2/20=8.1(O_{\text{Jan}} - E_{\text{Jan}})^2 / E_{\text{Jan}} = (11 - 20)^2 / 20 = 8.1
step 11
For February, (OFeb−EFeb)2/EFeb=(16−16)2/16=0(O_{\text{Feb}} - E_{\text{Feb}})^2 / E_{\text{Feb}} = (16 - 16)^2 / 16 = 0
step 12
For March, (OMar−EMar)2/EMar=(18−16)2/16=0.25(O_{\text{Mar}} - E_{\text{Mar}})^2 / E_{\text{Mar}} = (18 - 16)^2 / 16 = 0.25
step 13
For April, (OApr−EApr)2/EApr=(19−12)2/12=4.083(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: χ2=0+8.1+0+0.25+4.083=12.433\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=4df = 5 - 1 = 4
step 16
Use a chi-square distribution table or calculator to find the P-value corresponding to χ2=12.433\chi^2 = 12.433 and df=4df = 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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