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:
-
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: \[ {-0.72, -1.78, 0.31, 1.09, 3.76, 1.45, 2.10, 0.01, -0.06, 1.00} \]
-
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 \( s_d \): \[ s_d = \sqrt{\frac{21.4085}{10-1}} = \sqrt{\frac{21.4085}{9}} \approx \sqrt{2.3765} \approx 1.545 \]
-
-
Calculate the t-statistic: \[ t = \frac{\bar{d}}{s_d/\sqrt{n}} = \frac{0.715}{1.545/\sqrt{10}} \approx \frac{0.715}{0.488} \approx 1.464 \]
-
Determine the Degrees of Freedom: \[ df = n - 1 = 10 - 1 = 9 \]
-
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:
\[ \text{P-value} \approx 0.1775 \]