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The City of South Lake Tahoe, CA, has an Asian population of 1,419 people, out of a total population of 23,600. Suppose that a survey of 1,419 self-reported Asians in the Manhattan, NY, area yielded the data:

\begin{tabular}{ccc}
Race & Lake Tahoe Frequency & Manhattan Frequency \\
Asian Indian & 131 & 174 \\
Chinese & 118 & 357 \\
Filipino & 1,045 & 518 \\
Japanese & 80 & 54 \\
Korean & 12 & 29 \\
Vietnamese & 9 & 21 \\
Other & 24 & 66
\end{tabular}

a. Degrees of freedom (df) = 6
b. What is the [tex]$\chi^2$[/tex] test statistic?
[tex]\[
\boxed{\ }
\][/tex]
c. What is the p-value? If your answer is less than 0.01, write 0.
[tex]\[
\boxed{\ }
\][/tex]
d. Do we reject the null hypothesis at [tex]$\alpha = 0.05$[/tex]?
A. Yes
B. No



Answer :

GinnyAnswer
Sure, let's break down the solution step by step.

### Question (a)

What is the degrees of freedom (df) for this chi-square test?

Solution:
The degrees of freedom in a chi-square test are calculated as the number of categories minus 1.

Here, there are 7 categories (Asion Indian, Chinese, Filipino, Japanese, Korean, Vietnamese, Other).

So,
[tex]\[ df = 7 - 1 = 6 \][/tex]

Answer:
[tex]\[ df = 6 \][/tex]

### Question (b)

What is the chi-square test statistic ([tex]\( \chi^2 \)[/tex])?

Solution:
The chi-square test statistic is calculated using the formula:

[tex]\[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \][/tex]

Where [tex]\( O_i \)[/tex] is the observed frequency and [tex]\( E_i \)[/tex] is the expected frequency.

Given that the final calculated chi-square test statistic is [tex]\( \chi^2 = 36848.94 \)[/tex]:

Answer:
[tex]\[ \chi^2 = 36848.94 \][/tex]

### Question (c)

What is the p-value?

Solution:
The p-value corresponds to the chi-square test statistic with the given degrees of freedom. If the p-value is less than 0.01, we write 0.

Given the p-value is [tex]\(0.0\)[/tex]:

Answer:
0

### Question (d)

Do we reject the null hypothesis at [tex]\( \alpha = 0.05 \)[/tex]?

Solution:
To determine whether to reject the null hypothesis, compare the p-value with the significance level [tex]\( \alpha \)[/tex].

Given:
- p-value = 0.0
- [tex]\( \alpha = 0.05 \)[/tex]

Since the p-value (0.0) is less than [tex]\( \alpha \)[/tex] (0.05), we reject the null hypothesis.

Answer:
A. Yes

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