To conduct the hypothesis test, we will set up the null and alternative hypotheses based on the problem statement.
Step 1: Define the Hypotheses
Let \( uvd \) represent the mean increase in height between ages 12 and 14.
-
Null Hypothesis (\( H_0 \)): The mean increase in height is at least 5 inches. \[ H_0: uvd \geq 5 \]
-
Alternative Hypothesis (\( H_a \): The mean increase in height is less than 5 inches. \[ H_a: uvd < 5 \]
Step 2: Collect the Data
Heights at Age 12 and Age 14
-
Age 12:
- 60.7, 55.9, 59.7, 56.3, 63.0, 58.6, 61.1, 59.5, 61.6, 59.3
-
Age 14:
- 65.5, 62.1, 65.4, 61.5, 68.5, 63.9, 65.8, 64.5, 66.3, 64.2
Calculate the Increases
We calculate the differences \( d_i = \text{Height at Age 14} - \text{Height at Age 12} \).
\[ \begin{align*} d_1 & = 65.5 - 60.7 = 4.8 \ d_2 & = 62.1 - 55.9 = 6.2 \ d_3 & = 65.4 - 59.7 = 5.7 \ d_4 & = 61.5 - 56.3 = 5.2 \ d_5 & = 68.5 - 63.0 = 5.5 \ d_6 & = 63.9 - 58.6 = 5.3 \ d_7 & = 65.8 - 61.1 = 4.7 \ d_8 & = 64.5 - 59.5 = 5.0 \ d_9 & = 66.3 - 61.6 = 4.7 \ d_{10} & = 64.2 - 59.3 = 4.9 \ \end{align*} \]
- Differences: \( d = [4.8, 6.2, 5.7, 5.2, 5.5, 5.3, 4.7, 5.0, 4.7, 4.9] \)
Step 3: Summary Statistics
Calculate the sample mean \( \bar{d} \) and the sample standard deviation \( s_d \).
Mean Increase \( \bar{d} \)
\[ \bar{d} = \frac{4.8 + 6.2 + 5.7 + 5.2 + 5.5 + 5.3 + 4.7 + 5.0 + 4.7 + 4.9}{10} = \frac{57.0}{10} = 5.7 \]
Standard Deviation \( s_d \)
\[ s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n - 1}} \] Calculating each squared difference: \[ \begin{align*} (4.8 - 5.57)^2 & = 0.5921 \ (6.2 - 5.57)^2 & = 0.3969 \ (5.7 - 5.57)^2 & = 0.0169 \ (5.2 - 5.57)^2 & = 0.1369 \ (5.5 - 5.57)^2 & = 0.0049 \ (5.3 - 5.57)^2 & = 0.0729 \ (4.7 - 5.57)^2 & = 0.7569 \ (5.0 - 5.57)^2 & = 0.3249 \ (4.7 - 5.57)^2 & = 0.7569 \ (4.9 - 5.57)^2 & = 0.4489 \ \end{align*} \]
Calculating the sum of squared differences: \[ \sum (d_i - \bar{d})^2 = 0.5921 + 0.3969 + 0.0169 + 0.1369 + 0.0049 + 0.0729 + 0.7569 + 0.3249 + 0.7569 + 0.4489 = 2.6513 \]
Now we can find \( s_d \): \[ s_d = \sqrt{\frac{2.6513}{9}} \approx \sqrt{0.2946} \approx 0.543 \]
Step 4: Test Statistic
Using the t-statistic: \[ t = \frac{\bar{d} - 5}{s_d/\sqrt{n}} \] Where \( n = 10 \): \[ t = \frac{5.7 - 5}{0.543/\sqrt{10}} = \frac{0.7}{0.171} \approx 4.09 \]
Step 5: Determine the P-Value
Now we will use the t-distribution with \( n - 1 = 9 \) degrees of freedom to find the p-value for our calculated t-value.
Using a t-table or calculator:
- Degrees of freedom: 9
- Calculated t-value: 4.09
We look up or calculate the p-value for \( t = 4.09 \). This would be very small, indicating strong evidence against the null hypothesis.
Step 6: Conclusion
Since we are using \( \alpha = 0.01 \):
- If \( P \) value < 0.01, reject \( H_0 \).
Given that the calculated t-value is significant and the resulting p-value is very small, we reject the null hypothesis that the mean increase in height is at least 5 inches.
Final Conclusion:
There is sufficient evidence at the \( \alpha = 0.01 \) level to conclude that the mean increase in height is less than 5 inches.
4 answers