Question 1171313: Inference (Two Populations), Chi-Squared Tests
1. A study comparing children’s reading age (in months) was developed using identical twin
toddlers. One set of 6 twins played for 2 hours each day with educational toys
(experimental group), the corresponding set of 6 twins played for 2 hours each day with
non-educational toys (control group). The mean difference in reading age between the
experimental group and control group of 6 sets of twins was -2.33 months and the
standard deviation of the sample difference was 2.16 months. Set up a hypothesis test to
determine if there is a difference between the 2 groups and use the appropriate sample
test statistic to determine if the difference in reading age is significant at the 5% level
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Absolutely! Let's break down this hypothesis test step-by-step.
**1. Set up the Hypotheses:**
* **Null Hypothesis (H₀):** There is no difference in the mean reading age between the experimental group (educational toys) and the control group (non-educational toys). In other words, the mean difference is zero.
* H₀: μd = 0 (where μd is the mean difference)
* **Alternative Hypothesis (H₁):** There is a difference in the mean reading age between the two groups.
* H₁: μd ≠ 0 (two-tailed test)
**2. Determine the Test Statistic:**
Since we are comparing the means of two related groups (identical twins) and we have a small sample size (n = 6), we will use a paired t-test.
The formula for the t-statistic is:
t = (mean difference) / (standard deviation of the difference / √sample size)
t = μd / (sd / √n)
Where:
* μd = -2.33 months (mean difference)
* sd = 2.16 months (standard deviation of the difference)
* n = 6 (number of twin pairs)
**3. Calculate the Test Statistic:**
t = -2.33 / (2.16 / √6)
t = -2.33 / (2.16 / 2.449)
t = -2.33 / 0.882
t ≈ -2.642
**4. Determine the Degrees of Freedom:**
Degrees of freedom (df) = n - 1
df = 6 - 1 = 5
**5. Determine the Critical Value or P-value:**
* **P-value Approach:**
* Using a t-distribution table or a calculator, we find the p-value associated with a t-statistic of -2.642 and 5 degrees of freedom.
* Since this is a two-tailed test, we look for the probability of observing a t-statistic as extreme as -2.642 or 2.642.
* The p-value is approximately 0.0459.
* **Critical Value Approach:**
* For a two-tailed test with α = 0.05 and df = 5, the critical t-values are ±2.571.
**6. Make a Decision:**
* **P-value Approach:**
* Since the p-value (0.0459) is less than the significance level (0.05), we reject the null hypothesis.
* **Critical Value Approach:**
* Since the calculated t-statistic (-2.642) is less than the critical value (-2.571), we reject the null hypothesis.
**7. Conclusion:**
There is sufficient evidence to conclude that there is a statistically significant difference in the mean reading age between children who play with educational toys and those who play with non-educational toys at the 5% significance level.
|
|
|
