Question

In an experiment to determine whether there is a systematic difference between the weights obtained with two different scales, 10 rock specimens were weighed, in grams, on each scale. The following data were obtained:
Specimen 1 : weight on scale 1: 12.13 , weight on scale 2: 12.85
Specimen 2: weight on scale 1: 15.51, weight on scale 2: 17.29
specimen 3: 9.83 / 9.52
specimen 4: 12.39/11.30
specimen 5: 27.64/23.88
specimen 6: 10.80/9.35
specimen 7: 15.89/13.79
specimen 8: 6.99/6.98
specimen 9: 13.39/13.45
specimen 10: 21.90/20.9

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

Answer 1

P-value is 0.0885.

Do we reject Hv0?

Answer 2

There [is or isnt?] enough evidence to conclude that the the mean weight on Scale 1 is greater than the mean weight on Scale 2.

Answer 3

To determine if there is a systematic difference between the weights obtained from the two scales, we can perform a paired t-test. The steps are as follows:

1. **Calculate the Differences (d)**:
We find the differences between the weights on Scale 1 and Scale 2 for each specimen:

\[
d = \text{(Weight on Scale 1)} - \text{(Weight on Scale 2)}
\]

The data is as follows:

\[
\begin{align*}
\text{Specimen 1: } & 12.13 - 12.85 = -0.72 \\
\text{Specimen 2: } & 15.51 - 17.29 = -1.78 \\
\text{Specimen 3: } & 9.83 - 9.52 = 0.31 \\
\text{Specimen 4: } & 12.39 - 11.30 = 1.09 \\
\text{Specimen 5: } & 27.64 - 23.88 = 3.76 \\
\text{Specimen 6: } & 10.80 - 9.35 = 1.45 \\
\text{Specimen 7: } & 15.89 - 13.79 = 2.10 \\
\text{Specimen 8: } & 6.99 - 6.98 = 0.01 \\
\text{Specimen 9: } & 13.39 - 13.45 = -0.06 \\
\text{Specimen 10: } & 21.90 - 20.90 = 1.00 \\
\end{align*}
\]

The differences \( d \) are:
\[
\{-0.72, -1.78, 0.31, 1.09, 3.76, 1.45, 2.10, 0.01, -0.06, 1.00\}
\]

2. **Calculate the Mean and Standard Deviation of the Differences**:

- Mean \( \bar{d} \):
\[
\bar{d} = \frac{\sum d_i}{n} = \frac{-0.72 - 1.78 + 0.31 + 1.09 + 3.76 + 1.45 + 2.10 + 0.01 - 0.06 + 1.00}{10} = \frac{7.15}{10} = 0.715
\]

- Standard Deviation \( s_d \):
\[
s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}}
\]

First, we need to calculate the squared deviations:
\[
\begin{align*}
& (-0.72 - 0.715)^2 = (-1.435)^2 = 2.058225 \\
& (-1.78 - 0.715)^2 = (-2.495)^2 = 6.223025 \\
& (0.31 - 0.715)^2 = (-0.405)^2 = 0.164025 \\
& (1.09 - 0.715)^2 = (0.375)^2 = 0.140625 \\
& (3.76 - 0.715)^2 = (3.045)^2 = 9.273025 \\
& (1.45 - 0.715)^2 = (0.735)^2 = 0.540225 \\
& (2.10 - 0.715)^2 = (1.385)^2 = 1.920225 \\
& (0.01 - 0.715)^2 = (-0.705)^2 = 0.497025 \\
& (-0.06 - 0.715)^2 = (-0.775)^2 = 0.600625 \\
& (1.00 - 0.715)^2 = (0.285)^2 = 0.081225 \\
\end{align*}
\]

Now, summing these values gives us:
\[
\sum (d_i - \bar{d})^2 = 2.058225 + 6.223025 + 0.164025 + 0.140625 + 9.273025 + 0.540225 + 1.920225 + 0.497025 + 0.600625 + 0.081225 = 21.4085
\]

Now plug in to find \( s_d \):
\[
s_d = \sqrt{\frac{21.4085}{10-1}} = \sqrt{\frac{21.4085}{9}} \approx \sqrt{2.3765} \approx 1.545
\]

3. **Calculate the t-statistic**:
\[
t = \frac{\bar{d}}{s_d/\sqrt{n}} = \frac{0.715}{1.545/\sqrt{10}} \approx \frac{0.715}{0.488} \approx 1.464
\]

4. **Determine the Degrees of Freedom**:
\[
df = n - 1 = 10 - 1 = 9
\]

5. **Find the P-value**:
You can use a t-distribution table or calculator to find the P-value for \( t \approx 1.464 \) with \( df = 9 \). Performing a two-tailed test:

Using statistical software or a calculator, the P-value can be obtained. For \( t(9) = 1.464 \):

After looking up the value in a t-distribution table or using software:

The P-value ≈ 0.1775.

Thus, the P-value to four decimal places is:

\[
\text{P-value} \approx 0.1775
\]

Answer 4

To determine whether to reject the null hypothesis (\(H_0\)), we need to compare the P-value to a significance level (\(\alpha\)). The most common significance levels are 0.05 and 0.01.

### Given:
- **P-value**: 0.0885

### Decision Rule:
- If \( \text{P-value} < \alpha \), reject \(H_0\).
- If \( \text{P-value} \geq \alpha \), do not reject \(H_0\).

### Considerations:
1. **If \(\alpha = 0.05\)**:
- Since \(0.0885 > 0.05\), we **do not reject \(H_0\)**.

2. **If \(\alpha = 0.01\)**:
- Since \(0.0885 > 0.01\), we **do not reject \(H_0\)**.

### Conclusion:
In both scenarios, we do not have sufficient evidence to reject the null hypothesis (\(H_0\)). This implies that there is no statistically significant difference in weights between the two scales at the commonly used significance levels.

Answer 5

Given that the P-value (0.0885) is greater than the common significance levels (such as 0.05 and 0.01), we do not reject the null hypothesis (\(H_0\)).

### Conclusion:
There **isn't** enough evidence to conclude that the mean weight on Scale 1 is greater than the mean weight on Scale 2.

This means that we do not have sufficient statistical support to claim that there is a systematic difference in favor of Scale 1 having a higher mean weight compared to Scale 2.