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Math  /  Data & Statistics

QuestionGiven the following data in measuring a distance of a certain line. \begin{tabular}{|l|l|} \hline DISTANCE & NO. OF MEASUREMENTS \\ \hline 740.53 & 4 \\ \hline 740.59 & 3 \\ \hline 740.57 & 6 \\ \hline 740.53 & 7 \\ \hline \end{tabular}
Calculate the standard error of the mean (A) ±0.0037\pm 0.0037 (B) ±0.0247\pm 0.0247 (C) ±0.0055\pm 0.0055 (D) =0.0167=0.0167

Studdy Solution

STEP 1

1. The data provided consists of different distance measurements and the corresponding number of times each measurement was taken.
2. The standard error of the mean (SEM) is calculated using the formula: SEM=sn \text{SEM} = \frac{s}{\sqrt{n}} where s s is the standard deviation of the sample, and n n is the total number of measurements.
3. To find s s , we need to calculate the mean of the measurements, the deviations from the mean, the squared deviations, and then the variance.

STEP 2

1. Calculate the mean of the measurements.
2. Calculate the deviations from the mean for each measurement.
3. Square the deviations and find the variance.
4. Calculate the standard deviation from the variance.
5. Use the standard deviation and the total number of measurements to find the standard error of the mean (SEM).

STEP 3

Calculate the total number of measurements, n n .
n=4+3+6+7=20 n = 4 + 3 + 6 + 7 = 20

STEP 4

Calculate the weighted sum of the measurements.
Weighted sum=4×740.53+3×740.59+6×740.57+7×740.53 \text{Weighted sum} = 4 \times 740.53 + 3 \times 740.59 + 6 \times 740.57 + 7 \times 740.53
=2962.12+2221.77+4443.42+5183.71=14810.02 = 2962.12 + 2221.77 + 4443.42 + 5183.71 = 14810.02

STEP 5

Calculate the mean of the measurements.
xˉ=Weighted sumn=14810.0220=740.501 \bar{x} = \frac{\text{Weighted sum}}{n} = \frac{14810.02}{20} = 740.501

STEP 6

Calculate the deviations from the mean for each measurement.
For 740.53: 740.53−740.501=0.029 740.53 - 740.501 = 0.029
For 740.59: 740.59−740.501=0.089 740.59 - 740.501 = 0.089
For 740.57: 740.57−740.501=0.069 740.57 - 740.501 = 0.069
For 740.53: 740.53−740.501=0.029 740.53 - 740.501 = 0.029

STEP 7

Square the deviations and calculate the weighted sum of the squared deviations.
For 740.53 (first instance): 4×(0.029)2=4×0.000841=0.003364 4 \times (0.029)^2 = 4 \times 0.000841 = 0.003364
For 740.59: 3×(0.089)2=3×0.007921=0.023763 3 \times (0.089)^2 = 3 \times 0.007921 = 0.023763
For 740.57: 6×(0.069)2=6×0.004761=0.028566 6 \times (0.069)^2 = 6 \times 0.004761 = 0.028566
For 740.53 (second instance): 7×(0.029)2=7×0.000841=0.005887 7 \times (0.029)^2 = 7 \times 0.000841 = 0.005887
Total weighted sum of squared deviations: 0.003364+0.023763+0.028566+0.005887=0.06158 0.003364 + 0.023763 + 0.028566 + 0.005887 = 0.06158

STEP 8

Calculate the variance s2 s^2 .
s2=Total weighted sum of squared deviationsn−1=0.0615819≈0.00324 s^2 = \frac{\text{Total weighted sum of squared deviations}}{n-1} = \frac{0.06158}{19} \approx 0.00324

STEP 9

Calculate the standard deviation s s .
s=s2=0.00324≈0.057 s = \sqrt{s^2} = \sqrt{0.00324} \approx 0.057

STEP 10

Calculate the standard error of the mean (SEM).
SEM=sn=0.05720≈0.0574.472≈0.0127 \text{SEM} = \frac{s}{\sqrt{n}} = \frac{0.057}{\sqrt{20}} \approx \frac{0.057}{4.472} \approx 0.0127
Solution: None of the provided options exactly match the calculated SEM of approximately ±0.0127 \pm 0.0127 . Therefore, it seems there might be an error in the provided options or possibly a miscalculation in the steps. Re-evaluating might be necessary.

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