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Observed Frequencies
\begin{tabular}{ccccc}
\hline Over 65 years & [tex]$45-65$[/tex] years & [tex]$25-44$[/tex] years & [tex]$15-24$[/tex] years & Under 15 years \\
\hline 94 & 282 & 549 & 235 & 408 \\
\hline
\end{tabular}

You wonder if the age distribution of people who are left-handed matches the age distribution of the US population. You obtain the following data from the 2000 census:
\begin{tabular}{|c|c|c|c|c|}
\hline \multicolumn{5}{|c|}{Percent Distribution of the US Population by Age Group} \\
\hline Age Group & Over 65 years & [tex]$45-65$[/tex] years & [tex]$25-44$[/tex] years & [tex]$15-24$[/tex] years & Under 15 years \\
\hline Percent & [tex]$12.43 \%$[/tex] & [tex]$22.01 \%$[/tex] & [tex]$30.22 \%$[/tex] & [tex]$13.93 \%$[/tex] & [tex]$21.41 \%$[/tex] \\
\hline
\end{tabular}

[Source: Hobbs, F., & Stoops, N. (2002). Census 2000 special reports: Demographic trends in the 20th century. US Census Bureau.]

You use a chi-square test for goodness of fit to see how well the sample of people who are left-handed fits the census data.

What is the most appropriate null hypothesis?

A. The distribution of ages among people who are left-handed is different from that provided by the census data.
B. The distribution of ages among people who are left-handed is equal across the five age categories.
C. The distribution of ages among people who are left-handed is not equal across the five age categories.
D. The distribution of ages among people who are left-handed is the same as that provided by the census data.

Fill in the missing values in the following table indicating the expected frequencies in each category for a sample size of 1,568, if the null hypothesis is true:

Expected Frequencies
\begin{tabular}{ccccc}
\hline Over 65 years & [tex]$45-65$[/tex] years & [tex]$25-44$[/tex] years & [tex]$15-24$[/tex] years & Under 15 years \\
\hline & & & & & \\
\hline
\end{tabular}


Sagot :

To determine the expected frequencies for a sample size of 1,568 based on the provided percent distribution of the US population by age group, we follow these steps:

1. Identify the Sample Size and Percent Distribution:
- Sample Size ([tex]\( n \)[/tex]) = 1,568
- Percent distribution:
- Over 65 years: [tex]\( 12.43\% \)[/tex]
- 45-65 years: [tex]\( 22.01\% \)[/tex]
- 25-44 years: [tex]\( 30.22\% \)[/tex]
- 15-24 years: [tex]\( 13.93\% \)[/tex]
- Under 15 years: [tex]\( 21.41\% \)[/tex]

2. Convert Percentages to Proportions:
- Convert each percentage into a proportion by dividing by 100.

For "Over 65 years":
[tex]\[ 12.43\% = 0.1243 \][/tex]

For "45-65 years":
[tex]\[ 22.01\% = 0.2201 \][/tex]

For "25-44 years":
[tex]\[ 30.22\% = 0.3022 \][/tex]

For "15-24 years":
[tex]\[ 13.93\% = 0.1393 \][/tex]

For "Under 15 years":
[tex]\[ 21.41\% = 0.2141 \][/tex]

3. Calculate the Expected Frequencies:
- Multiply each proportion by the sample size ([tex]\( n = 1,568 \)[/tex]) to obtain the expected frequency for each age group:

For "Over 65 years":
[tex]\[ 0.1243 \times 1568 = 194.9024 \][/tex]

For "45-65 years":
[tex]\[ 0.2201 \times 1568 = 345.1168 \][/tex]

For "25-44 years":
[tex]\[ 0.3022 \times 1568 = 473.8496 \][/tex]

For "15-24 years":
[tex]\[ 0.1393 \times 1568 = 218.4224 \][/tex]

For "Under 15 years":
[tex]\[ 0.2141 \times 1568 = 335.7088 \][/tex]

4. Present the Expected Frequencies:
We can now fill in the expected frequencies in the table:

[tex]\[ \begin{array}{|c|c|c|c|c|} \hline \text{Age Group} & \text{Expected Frequency} \\ \hline \text{Over 65 years} & 194.9024 \\ \hline \text{45-65 years} & 345.1168 \\ \hline \text{25-44 years} & 473.8496 \\ \hline \text{15-24 years} & 218.4224 \\ \hline \text{Under 15 years} & 335.7088 \\ \hline \end{array} \][/tex]

5. Formulating the Null Hypothesis:
The most appropriate null hypothesis (H₀) based on the data comparison would be:
[tex]\[ \text{The distribution of ages among people who are left-handed is the same as that provided by the census data.} \][/tex]
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