Question

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
Height age 12: 61.1, 62.7, 67.5, 59.2, 61.4, 60.7, 55.9, 59.7, 56.3, 63.0
height age 14: 65.9, 67.5, 73.7, 64.9, 67.0, 65.5, 61.1, 65.4, 61.5, 68.5

Hv0 :[?]
Hv1: [?]

This is a [left/right or two tailed?] test?

Answer 1

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

Answer 2

To conduct this hypothesis test, we first need to set up our null and alternative hypotheses based on the information provided.

### Null and Alternative Hypotheses

1. **Null Hypothesis (H0)**: The mean increase in height for the boys is equal to or greater than 5.6 inches. Symbolically, we can state this as:
\[
H_0: \mu_d \geq 5.6
\]
where \(\mu_d\) is the mean change in height from age 12 to age 14.

2. **Alternative Hypothesis (H1)**: The mean increase in height for the boys is less than 5.6 inches. Symbolically, we can state this as:
\[
H_1: \mu_d < 5.6
\]

### Type of Test

Since we are testing if the mean increase is less than a specific value (5.6 inches), this is a **left-tailed test**.

### Calculating the Mean Increase in Height

We need to calculate the differences in height for each boy at age 14 minus their height at age 12:

- Differences \(d_i\) are calculated as follows:
\[
d_1 = 65.9 - 61.1 = 4.8
\]
\[
d_2 = 67.5 - 62.7 = 4.8
\]
\[
d_3 = 73.7 - 67.5 = 6.2
\]
\[
d_4 = 64.9 - 59.2 = 5.7
\]
\[
d_5 = 67.0 - 61.4 = 5.6
\]
\[
d_6 = 65.5 - 60.7 = 4.8
\]
\[
d_7 = 61.1 - 55.9 = 5.2
\]
\[
d_8 = 65.4 - 59.7 = 5.7
\]
\[
d_9 = 61.5 - 56.3 = 5.2
\]
\[
d_{10} = 68.5 - 63.0 = 5.5
\]

Now we can summarize these differences:
\[
d = [4.8, 4.8, 6.2, 5.7, 5.6, 4.8, 5.2, 5.7, 5.2, 5.5]
\]

### Mean and Standard Deviation of Differences

Now we calculate the mean and standard deviation of \(d\):

- **Mean of differences** \(\bar{d}\):
\[
\bar{d} = \frac{\sum d_i}{n} = \frac{4.8 + 4.8 + 6.2 + 5.7 + 5.6 + 4.8 + 5.2 + 5.7 + 5.2 + 5.5}{10} = \frac{58.3}{10} = 5.83
\]

- **Standard deviation of the differences** \((s_d)\):
Calculating the standard deviation:
\[
s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n - 1}}
\]
Calculating each squared deviation:
1. \((4.8 - 5.83)^2\)
2. \((4.8 - 5.83)^2\)
3. \((6.2 - 5.83)^2\)
4. \((5.7 - 5.83)^2\)
5. \((5.6 - 5.83)^2\)
6. \((4.8 - 5.83)^2\)
7. \((5.2 - 5.83)^2\)
8. \((5.7 - 5.83)^2\)
9. \((5.2 - 5.83)^2\)
10. \((5.5 - 5.83)^2\)

Calculating these values and then dividing by \(n - 1 = 9\) will give you the standard deviation.

### Test Statistic

Then, we calculate the t-test statistic using:
\[
t = \frac{\bar{d} - \mu_0}{s_d/\sqrt{n}}
\]
where \(\mu_0 = 5.6\).

### P-value and Conclusion

We'll find the p-value associated with this t-value using a t-distribution with \(n - 1\) degrees of freedom and compare the p-value to the significance level of \(0.01\).

If the p-value is less than or equal to \(0.01\), we reject the null hypothesis, concluding that the mean increase in height is less than 5.6 inches. If not, we fail to reject the null hypothesis.

This overall framework provides a systematic approach to concluding whether the mean increase in height is statistically less than 5.6 inches.

Please proceed with calculation based on this framework.

Answer 3

To compute the P-value, let's follow the outlined steps and perform the necessary calculations.

### Step 1: Calculate the Differences

We already calculated the height differences from age 12 to age 14 as follows:

\[
d = [4.8, 4.8, 6.2, 5.7, 5.6, 4.8, 5.2, 5.7, 5.2, 5.5]
\]

### Step 2: Calculate the Mean of the Differences

The mean of the differences (\(\bar{d}\)) has been computed as:
\[
\bar{d} = 5.83
\]

### Step 3: Calculate the Standard Deviation of the Differences

To calculate the standard deviation (\(s_d\)), follow these steps:

1. Calculate the squared deviations from mean:
- \( (4.8 - 5.83)^2 = (-1.03)^2 = 1.0609 \)
- \( (4.8 - 5.83)^2 = (-1.03)^2 = 1.0609 \)
- \( (6.2 - 5.83)^2 = (0.37)^2 = 0.1369 \)
- \( (5.7 - 5.83)^2 = (-0.13)^2 = 0.0169 \)
- \( (5.6 - 5.83)^2 = (-0.23)^2 = 0.0529 \)
- \( (4.8 - 5.83)^2 = (-1.03)^2 = 1.0609 \)
- \( (5.2 - 5.83)^2 = (-0.63)^2 = 0.3969 \)
- \( (5.7 - 5.83)^2 = (-0.13)^2 = 0.0169 \)
- \( (5.2 - 5.83)^2 = (-0.63)^2 = 0.3969 \)
- \( (5.5 - 5.83)^2 = (-0.33)^2 = 0.1089 \)

2. Sum the squared deviations:
\[
1.0609 + 1.0609 + 0.1369 + 0.0169 + 0.0529 + 1.0609 + 0.3969 + 0.0169 + 0.3969 + 0.1089 = 4.722
\]

3. Calculate the variance (\(s_d^2\)):
\[
s_d^2 = \frac{4.722}{n - 1} = \frac{4.722}{9} \approx 0.5247
\]

4. Calculate the standard deviation (\(s_d\)):
\[
s_d = \sqrt{0.5247} \approx 0.7240
\]

### Step 4: Calculate the Test Statistic

Now, we calculate the t-statistic using:
\[
t = \frac{\bar{d} - \mu_0}{s_d/\sqrt{n}}
\]
where \(\mu_0 = 5.6\) and \(n = 10\). Thus:
\[
t = \frac{5.83 - 5.6}{0.7240/\sqrt{10}}
\]
\[
t = \frac{0.23}{0.7240/\sqrt{10}} = \frac{0.23}{0.7240/3.1623} \approx \frac{0.23}{0.2297} \approx 1.000
\]

### Step 5: Calculate the P-value

Since this is a left-tailed test, we need to find the P-value for \(t = 1.000\) with \(n - 1 = 9\) degrees of freedom.

Using a t-distribution table or a calculator, we can find:
\[
P(T < 1.000 \text{ for } df = 9)
\]

Using a t-table or calculator, we find that the P-value corresponds to:
\[
P(T < 1.000) \approx 0.8413
\]
Thus, the P-value for our test statistic is approximately \(0.8413\).

### Conclusion

The rounded P-value (to at least four decimal places) is:
\[
\text{P-value} \approx 0.8413
\]

### Final Note

Since the P-value is significantly greater than the significance level \(0.01\), we would not reject the null hypothesis.