To determine whether the external condition of temperature is affecting the growth of children's height at a 5% level of significance, we can use a hypothesis test. Since we are comparing the heights from two different locations (A and B), we can perform a two-sample t-test for the means.

Steps to Perform a Two-Sample t-Test:

1. State the Hypotheses:

   • Null Hypothesis (H₀): There is no significant difference between the mean heights of children from locations A and B (i.e., temperature does not affect growth). $H₀: \mu_A = \mu_B$
   • Alternative Hypothesis (H₁): There is a significant difference between the mean heights of children from locations A and B (i.e., temperature affects growth). $H₁: \mu_A \neq \mu_B$

2. Calculate the Means and Standard Deviations:

   • Heights from location A: 98, 102, 96, 104, 100
   • Heights from location B: 100, 105, 98, 102, 104

   We will compute the means ($\bar{x}_A$ and $\bar{x}_B$) and the standard deviations ($s_A$ and $s_B$) for both samples.

3. Determine the Degrees of Freedom and Compute the t-Statistic using the formula: $t = \frac{\bar{x}_A - \bar{x}_B}{\sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}}$ where:

   • $n_A$ and $n_B$ are the sample sizes for location A and B, respectively.
   • $s_A^2$ and $s_B^2$ are the variances (square of the standard deviations).

4. Find the Critical Value for a two-tailed test at the 5% level of significance.

5. Make a Decision:

   • If the computed t-value is greater than the critical value, reject the null hypothesis.
   • Otherwise, fail to reject the null hypothesis.

I'll now compute these values to perform the test.### Results of the Two-Sample t-Test:

1. Means:

   • Location A: Mean height = 100.0
   • Location B: Mean height = 101.8

2. Standard Deviations:

   • Location A: Standard deviation = 3.16
   • Location B: Standard deviation = 2.86

3. t-Statistic: $t = -0.943$

4. p-Value: $p = 0.373$

5. Critical t-Value: For a 5% significance level and degrees of freedom ≈ 7.92, the critical t-value is approximately $t_{\text{critical}} = 2.31$.

Conclusion:

• Since the absolute value of the computed t-statistic ($|t| = 0.943$) is less than the critical value (2.31), and the p-value (0.373) is greater than 0.05, we fail to reject the null hypothesis.
• Therefore, there is no statistically significant evidence to suggest that the external temperature conditions are affecting the growth of children's height at the 5% significance level.

Would you like further details on any of these steps or concepts?

Here are 5 related questions to explore further:

1. What assumptions are made in a two-sample t-test, and are they satisfied here?
2. How would the results change if the sample sizes were larger?
3. What is the difference between a paired and an unpaired t-test?
4. How would a one-tailed test differ from this two-tailed test?
5. Can other factors besides temperature affect the height difference in children?

Tip: Always check the sample size and variance before choosing a t-test, as it influences the assumptions about equal or unequal variances.