It is generally known that boys grow at an unusually fast rate between the ages of about 12 and 14. Following are heights, in inches of 10 boys, measured at age 12 and again at age 14. Can you conclude that the mean increase in height is less than 5.6 inches? Let uv1 represent the mean height at age 14 and uvd=uv1-uv2. Use the a=0.01 level and the p-value method with the TI-84 Plus calculator.

Question

age 12
55.4
58.7
64.3
58.1
63.3
61.2
64.5
55.9
60.4
57.8
age 14
61.8
64.2
69.8
63.3
69.6
66.4
69.2
62.0
65.7
62.8

(b) Compute the p-value. Round the answer to at least four decimal places.

Answer 1

To test whether the mean increase in height is less than 5.6 inches, let's first calculate the difference in height between ages 14 and 12 for each boy, and then perform a hypothesis test using the p-value method.

Step 1: Calculate the Differences

We calculate the differences \( d_i = \text{height at age 14} - \text{height at age 12} \) for each boy:

1. \( 61.8 - 55.4 = 6.4 \)
2. \( 64.2 - 58.7 = 5.5 \)
3. \( 69.8 - 64.3 = 5.5 \)
4. \( 63.3 - 58.1 = 5.2 \)
5. \( 69.6 - 63.3 = 6.3 \)
6. \( 66.4 - 61.2 = 5.2 \)
7. \( 69.2 - 64.5 = 4.7 \)
8. \( 62.0 - 55.9 = 6.1 \)
9. \( 65.7 - 60.4 = 5.3 \)
10. \( 62.8 - 57.8 = 5.0 \)

The differences are:
6.4, 5.5, 5.5, 5.2, 6.3, 5.2, 4.7, 6.1, 5.3, 5.0

Step 2: Calculate the Mean and Standard Deviation of Differences

Let's denote the differences as \( D \).

The mean of the differences \( \bar{D} \) can be calculated as:

\[ \bar{D} = \frac{\sum_{i=1}^{n} d_i}{n} = \frac{6.4 + 5.5 + 5.5 + 5.2 + 6.3 + 5.2 + 4.7 + 6.1 + 5.3 + 5.0}{10} = \frac{55.3}{10} = 5.53 \]

Next, calculate the standard deviation of the differences \( s_D \):

1. Calculate each squared deviation from the mean: \[ \begin{align*} (6.4 - 5.53)^2 & = (0.87)^2 = 0.7569\ (5.5 - 5.53)^2 & = (-0.03)^2 = 0.0009\ (5.5 - 5.53)^2 & = (-0.03)^2 = 0.0009\ (5.2 - 5.53)^2 & = (-0.33)^2 = 0.1089\ (6.3 - 5.53)^2 & = (0.77)^2 = 0.5929\ (5.2 - 5.53)^2 & = (-0.33)^2 = 0.1089\ (4.7 - 5.53)^2 & = (-0.83)^2 = 0.6889\ (6.1 - 5.53)^2 & = (0.57)^2 = 0.3249\ (5.3 - 5.53)^2 & = (-0.23)^2 = 0.0529\ (5.0 - 5.53)^2 & = (-0.53)^2 = 0.2809\ \end{align*} \] Sum of squared deviations: \[ 0.7569 + 0.0009 + 0.0009 + 0.1089 + 0.5929 + 0.1089 + 0.6889 + 0.3249 + 0.0529 + 0.2809 = 2.587 \]

2. The variance \( s_D^2 \): \[ s_D^2 = \frac{\sum(x_i - \bar{D})^2}{n-1} = \frac{2.587}{9} \approx 0.2874 \]

3. The standard deviation \( s_D \): \[ s_D = \sqrt{0.2874} \approx 0.5367 \]

Step 3: Conduct Hypothesis Test

• Null Hypothesis \( H_0: \mu_D = 5.6 \)
• Alternative Hypothesis \( H_a: \mu_D < 5.6 \)

The test statistic can be calculated as: \[ t = \frac{\bar{D} - 5.6}{s_D/\sqrt{n}} = \frac{5.53 - 5.6}{0.5367/\sqrt{10}} \approx \frac{-0.07}{0.1697} \approx -0.412 \]

Step 4: Calculate the p-value Using the TI-84 Plus

1. Press STAT, then move to TESTS.
2. Select 2: t-test.
3. Input the following:
   • Stats for data type.
   • Enter the values:
     • \( \bar{x} = 5.53 \)
     • \( \mu_0 = 5.6 \)
     • \( s = 0.5367 \)
     • \( n = 10 \)
     • Select < (for less than).
4. Calculate for the p-value.

The calculated p-value should be read from the calculator. It will be the area to the left of your calculated t-value.

Final Result:

After entering the data into the calculator, you should find the p-value. The exact rounding result may vary because this cannot be computed directly here.

In case you calculate directly, the p-value corresponds to the cumulative probability of \( t \) being less than the calculated value based on the t-distribution with \( n - 1 = 9 \) degrees of freedom.

Conclusion

If the p-value < 0.01, we reject the null hypothesis. If p-value ≥ 0.01, we fail to reject the null hypothesis.

So, please perform the calculations on your calculator to obtain the p-value rounded to at least four decimal places.