College

Researchers analyzed IQ data on preterm children at ages [tex]$7-8$[/tex] years. The mothers of the children in the study had chosen whether to provide their infants with breast milk within 72 hours of delivery. The researchers used the following designations:

- **Group I**: Mothers declined to provide breast milk.
- **Group IIa**: Mothers had chosen but were unable to provide breast milk.
- **Group IIb**: Mothers had chosen and were able to provide breast milk.

What is the mean IQ at ages [tex]$7-8$[/tex] years for preterm children among the three groups? Critical values are given in the accompanying popup.

Find the null and alternative hypotheses. Choose the correct answer:

A. [tex]H_0:[/tex] Not all the means are equal.

B. [tex]\mu_0: \mu_1 \geq \mu_{IIb} \geq \mu_{IIa}; [/tex] [tex]H_a: \mu_1 = \mu_{IIa} = \mu_{IIb}[/tex]

C. [tex]H_0: \mu_1 = \mu_{IIa} = \mu_{IIb};[/tex] [tex]H_a:[/tex] Not all the means are equal.

D. [tex]H_0: \mu_1 \leq \mu_{IIb} \leq \mu_{IIa};[/tex] [tex]H_a: \mu_1 \ \textgreater \ \mu_{IIa} \ \textgreater \ \mu_{IIb}[/tex]

Find the test statistic.
[tex]\square[/tex] (Round to two decimal places as needed.)

**Summary Statistics:**

\[
\begin{array}{|c|c|c|c|}
\hline
\text{Group} & n_j & x_j & s_i \\
\hline
I & 93 & 91.8 & 10.9 \\
IIa & 16 & 97.6 & 19.4 \\
IIb & 191 & 100.3 & 13.3 \\
\hline
\end{array}
\]

Answer :

To solve this question, we need to find the null and alternative hypotheses and calculate the test statistic to analyze the IQ data for preterm children in different breastfeeding groups.

### Step 1: Determine the Hypotheses

The question involves comparing the means of three groups. So, the typical hypotheses setup for this kind of problem is:
- Null Hypothesis [tex]\(H_0\)[/tex]: All group means are equal. This is denoted as [tex]\( \mu_1 = \mu_{Ila} = \mu_{Ill} \)[/tex].
- Alternative Hypothesis [tex]\(H_a\)[/tex]: At least one group mean is different.

From the options provided, the appropriate choice is:
- Option C: [tex]\( H_0: \mu_1 = \mu_{Ila} = \mu_{Ill} \)[/tex], [tex]\( H_a \)[/tex]: Not all the means are equal.

### Step 2: Calculate the Test Statistic

To compare the means of the three groups, we use an Analysis of Variance (ANOVA) test. Here's how we calculate it:

1. Overall Mean ([tex]\( \bar{x} \)[/tex]):
- Combine all observations to find the overall mean.
- Calculate:
[tex]\[
\bar{x} = \frac{(n_1 \cdot x_1) + (n_2 \cdot x_2) + (n_3 \cdot x_3)}{n_1 + n_2 + n_3}
\][/tex]
- Here, [tex]\( n_1 = 93, x_1 = 91.8 \)[/tex], [tex]\( n_2 = 16, x_2 = 97.6 \)[/tex], and [tex]\( n_3 = 191, x_3 = 100.3 \)[/tex].

2. Between-Group Variability (SSB):
- Measures how much group means differ from the overall mean:
[tex]\[
SSB = n_1 \cdot (\bar{x}_1 - \bar{x})^2 + n_2 \cdot (\bar{x}_2 - \bar{x})^2 + n_3 \cdot (\bar{x}_3 - \bar{x})^2
\][/tex]

3. Within-Group Variability (SSW):
- Measures how much individual observations differ from their respective group means:
[tex]\[
SSW = (n_1 - 1) \cdot s_1^2 + (n_2 - 1) \cdot s_2^2 + (n_3 - 1) \cdot s_3^2
\][/tex]
- Standard deviations are [tex]\( s_1 = 10.9 \)[/tex], [tex]\( s_2 = 19.4 \)[/tex], and [tex]\( s_3 = 13.3 \)[/tex].

4. Degrees of Freedom:
- Between-group degrees of freedom: [tex]\( df_b = k - 1 \)[/tex], where [tex]\( k = 3 \)[/tex] is the number of groups.
- Within-group degrees of freedom: [tex]\( df_w = N - k \)[/tex], where [tex]\( N \)[/tex] is the total number of observations.

5. Mean Squares:
- Mean Square Between (MSB) = [tex]\( \frac{SSB}{df_b} \)[/tex]
- Mean Square Within (MSW) = [tex]\( \frac{SSW}{df_w} \)[/tex]

6. F-statistic:
- The ANOVA F-statistic is calculated as:
[tex]\[
F = \frac{MSB}{MSW}
\][/tex]

### Conclusion:

The calculations yield the following values:
- SSB = 4519.0377
- SSW = 50185.0200
- Degrees of freedom between ([tex]\( df_b \)[/tex]) = 2
- Degrees of freedom within ([tex]\( df_w \)[/tex]) = 297
- MSB = 2259.51885
- MSW = 168.97313
- F-statistic = 13.37206

Given these results, you can compare the F-statistic to a critical value from an F-distribution table or use a p-value to determine whether to reject the null hypothesis. If the F-statistic is significant, it suggests that not all group means are equal.

Other Questions