Answer :
To solve this problem, let's go through it step-by-step using the information provided.
### Part a: Degrees of Freedom (df)
The degrees of freedom in a chi-square test for independence is calculated using the formula:
[tex]\[ \text{df} = (r - 1) \times (c - 1) \][/tex]
where [tex]\( r \)[/tex] is the number of rows and [tex]\( c \)[/tex] is the number of columns in the contingency table.
The provided table has 3 rows (Tahoe, Utah, Colorado) and 3 columns (not labeled but there are three data points in each row). So:
[tex]\[ \text{df} = (3 - 1) \times (3 - 1) = 2 \times 2 = 4 \][/tex]
So, the degrees of freedom is:
[tex]\[ \boxed{4} \][/tex]
### Part b: Chi-Square Test Statistic ([tex]\(\chi^2\)[/tex] statistic)
The chi-square test statistic provided is:
[tex]\[ \chi^2 \approx 10.5259 \][/tex]
So, the chi-square test statistic is:
[tex]\[ \boxed{10.5259} \][/tex]
### Part c: p-Value
The given p-value is:
[tex]\[ p \approx 0.0324 \][/tex]
As the p-value is greater than 0.01, we don't need to write 0.
So, the p-value is:
[tex]\[ \boxed{0.0324} \][/tex]
### Part d: Decision at [tex]\(\alpha = 0.05\)[/tex]
To decide whether to reject the null hypothesis, compare the p-value to the significance level ([tex]\(\alpha = 0.05\)[/tex]).
The rule is:
- If [tex]\( p < \alpha \)[/tex], reject the null hypothesis.
- If [tex]\( p \geq \alpha \)[/tex], do not reject the null hypothesis.
Given [tex]\( p \approx 0.0324 \)[/tex], which is less than [tex]\( 0.05 \)[/tex]:
[tex]\[ 0.0324 < 0.05 \][/tex]
Therefore, we reject the null hypothesis at [tex]\(\alpha = 0.05\)[/tex].
So, the answer is:
[tex]\[ \boxed{\rightarrow \odot \text{A. Yes}} \][/tex]
In summary:
- Degrees of freedom: [tex]\(\boxed{4}\)[/tex]
- Chi-square test statistic: [tex]\(\boxed{10.5259}\)[/tex]
- p-value: [tex]\(\boxed{0.0324}\)[/tex]
- Reject the null hypothesis at [tex]\(\alpha = 0.05\)[/tex]? [tex]\(\boxed{\rightarrow \odot \text{A. Yes}}\)[/tex]
### Part a: Degrees of Freedom (df)
The degrees of freedom in a chi-square test for independence is calculated using the formula:
[tex]\[ \text{df} = (r - 1) \times (c - 1) \][/tex]
where [tex]\( r \)[/tex] is the number of rows and [tex]\( c \)[/tex] is the number of columns in the contingency table.
The provided table has 3 rows (Tahoe, Utah, Colorado) and 3 columns (not labeled but there are three data points in each row). So:
[tex]\[ \text{df} = (3 - 1) \times (3 - 1) = 2 \times 2 = 4 \][/tex]
So, the degrees of freedom is:
[tex]\[ \boxed{4} \][/tex]
### Part b: Chi-Square Test Statistic ([tex]\(\chi^2\)[/tex] statistic)
The chi-square test statistic provided is:
[tex]\[ \chi^2 \approx 10.5259 \][/tex]
So, the chi-square test statistic is:
[tex]\[ \boxed{10.5259} \][/tex]
### Part c: p-Value
The given p-value is:
[tex]\[ p \approx 0.0324 \][/tex]
As the p-value is greater than 0.01, we don't need to write 0.
So, the p-value is:
[tex]\[ \boxed{0.0324} \][/tex]
### Part d: Decision at [tex]\(\alpha = 0.05\)[/tex]
To decide whether to reject the null hypothesis, compare the p-value to the significance level ([tex]\(\alpha = 0.05\)[/tex]).
The rule is:
- If [tex]\( p < \alpha \)[/tex], reject the null hypothesis.
- If [tex]\( p \geq \alpha \)[/tex], do not reject the null hypothesis.
Given [tex]\( p \approx 0.0324 \)[/tex], which is less than [tex]\( 0.05 \)[/tex]:
[tex]\[ 0.0324 < 0.05 \][/tex]
Therefore, we reject the null hypothesis at [tex]\(\alpha = 0.05\)[/tex].
So, the answer is:
[tex]\[ \boxed{\rightarrow \odot \text{A. Yes}} \][/tex]
In summary:
- Degrees of freedom: [tex]\(\boxed{4}\)[/tex]
- Chi-square test statistic: [tex]\(\boxed{10.5259}\)[/tex]
- p-value: [tex]\(\boxed{0.0324}\)[/tex]
- Reject the null hypothesis at [tex]\(\alpha = 0.05\)[/tex]? [tex]\(\boxed{\rightarrow \odot \text{A. Yes}}\)[/tex]
