91Ó°ÊÓ

a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume all assumptions are met. The Head Start Program provides a wide range of services to low-income children up to the age of 5 years and their families. Its goals are to provide services to improve social and learning skills and to improve health and nutrition status so that the participants can begin school on an equal footing with their more advantaged peers. The distribution of ages for participating children is as follows: \(4 \%\) five-year-olds, \(52 \%\) four-year-olds, \(34 \%\) threeyear-olds, and \(10 \%\) under 3 years. When the program was assessed in a particular region, it was found that of the 200 randomly selected participants, 20 were 5 years old, 120 were 4 years old, 40 were 3 years old, and 20 were under 3 years. Is there sufficient evidence at \(\alpha=\) 0.05 that the proportions differ from the program's? Use the \(P\) -value method.

Step by step solution

01

State the Hypotheses

The null hypothesis (H_0) and alternative hypothesis (H_1) need to be defined.- Null Hypothesis (H_0): The proportions of ages in the region follow the claimed distribution: \( p_1 = 0.04 \), \( p_2 = 0.52 \), \( p_3 = 0.34 \), \( p_4 = 0.10 \). - Alternative Hypothesis (H_1): At least one of the age proportions differs from the claimed distribution.

02

Identify each Claim

The claim to be tested is that the distribution of ages in the program is as follows: \( 4\% \) five-year-olds, \( 52\% \) four-year-olds, \( 34\% \) three-year-olds, and \( 10\% \) under 3 years.

03

Compute the Expected Frequencies

Calculate the expected frequencies by multiplying each proportion by the total number of participants (200):- Expected for 5-year-olds: \( 200 \times 0.04 = 8 \)- Expected for 4-year-olds: \( 200 \times 0.52 = 104 \)- Expected for 3-year-olds: \( 200 \times 0.34 = 68 \)- Expected for under 3 years: \( 200 \times 0.10 = 20 \)

04

Calculate the Test Statistic using Chi-Square Formula

The test statistic \( \chi^2 \) is calculated 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.- For 5-year-olds: \( \frac{(20 - 8)^2}{8} = 18 \)- For 4-year-olds: \( \frac{(120 - 104)^2}{104} = 2.46 \)- For 3-year-olds: \( \frac{(40 - 68)^2}{68} = 11.76 \)- For under 3 years: \( \frac{(20 - 20)^2}{20} = 0 \)So, \( \chi^2 = 18 + 2.46 + 11.76 + 0 = 32.22 \).

05

Find the Critical Value

For a given significance level \( \alpha = 0.05 \) and degrees of freedom \( df = k - 1 = 4 - 1 = 3 \) (where \( k \) is the number of categories), the critical value \( \chi^2_{0.05, 3} \) can be found using a chi-square distribution table. The critical value is approximately \( 7.81 \).

06

Make the Decision

The decision rule is: If the test statistic \( \chi^2 = 32.22 \) is greater than the critical value \( 7.81 \), reject the null hypothesis.Since \( 32.22 > 7.81 \), we reject the null hypothesis.

07

Summarize the Results

With \( \chi^2 = 32.22 \) and \( p < 0.05 \), there is sufficient evidence to conclude that the distribution of ages in the program differs significantly from the expected proportions. This suggests a deviation from the claimed age distribution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chi-Square Test

The Chi-Square Test is a statistical method used to determine if there is a significant difference between the expected frequencies and the observed frequencies in categorical data.
This test helps in understanding if the distribution of sample data matches an expected distribution. It's important in hypothesis testing and provides a framework to compare categorical variables.

• To conduct a Chi-Square Test, you first need your observed frequencies – these are the actual counts from your sample.
• Then, you calculate expected frequencies based on a specific proportional distribution, as claimed in your hypothesis.
• The formula used is \[\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\]where \(O_i\) is the observed frequency, and \(E_i\) is the expected frequency.

In this exercise, participants' ages were compared against expected age proportions.
Differences indicated by a significant result suggest that the observed age distribution does not align with the expected claims.

Null Hypothesis

The Null Hypothesis, often denoted as \(H_0\), is a statement that assumes there is no effect or no difference. It represents a default position that any kind of effect observed is due to chance or a null effect.
In the context of hypothesis testing, the null hypothesis embodies the specific claim you seek to test.

• In our exercise, \(H_0\) is that the age distribution in the tested region follows the same proportion as the claimed distribution.
• The null hypothesis is always paired with an alternative hypothesis (\(H_1\)), which suggests that there is an effect or a difference. Here, \(H_1\) notes that at least one of the age proportions differs from the expected claims.

Rejecting \(H_0\) implies evidence against the assumption that there are no differences, suggesting a potential discrepancy in the data beyond pure chance.

Critical Value

The Critical Value is a threshold that determines the cutoff point at which the null hypothesis is rejected.
It acts as a benchmark; if your test statistic exceeds this value, you have enough evidence to reject the null hypothesis.

• It's based on the chosen significance level \(\alpha\), which denotes the probability of rejecting a true null hypothesis.
• For a Chi-Square test, the critical value depends on your degrees of freedom, calculated as the number of categories minus one: \(df = k - 1\).
• In this exercise, with 4 categories and \(\alpha = 0.05\), our degrees of freedom is 3, producing a critical value of approximately 7.81.

If your calculated \(\chi^2\) is greater than the critical value, the null hypothesis is not supported by the observed data.

Significance Level

The Significance Level, denoted as \(\alpha\), is a predetermined threshold used in hypothesis testing to decide whether the null hypothesis should be rejected.
It's a measure of how much risk you're willing to take in incorrectly rejecting the null hypothesis.

• Common significance levels include 0.01, 0.05, and 0.10, with 0.05 being most common. These numbers represent the probability of a Type I error, or false positive.
• In this exercise, \(\alpha = 0.05\) implies that there's a 5% risk of concluding that a difference exists when there actually isn’t one.
• The significance level helps define the critical region of your test statistic and affects the critical value.

By deciding the significance level, you play a key role in controlling the sensitivity and specificity of your hypothesis testing outcomes.