Question at position 4 Suppose you enter the following values for vx,0v_{x,0} in Table 7.1: [table]
vx,0v_{x,0} (m/s) 6.34 6.53 6.43 6.48 6.58
[/table] For this data set, v¯x,0=6.47\bar{v}_{x,0}=6.47 m/s. What is the uncertainty in v¯x,0\bar{v}_{x,0}? ±0.01\pm 0.01 m/s±0.04\pm 0.04 m/s±0.06\pm 0.06 m/s±0.08\pm 0.08 m/s±0.09\pm 0.09 m/s±0.11\pm 0.11 m/s±0.15\pm 0.15 m/s±0.21\pm 0.21 m/sClear my selection

Question

Question at position 4  Suppose you enter the following values for vx,0v_{x,0} in Table 7.1: [table]
vx,0v_{x,0} (m/s) 6.34 6.53 6.43 6.48 6.58
[/table] For this data set, v¯x,0=6.47\bar{v}_{x,0}=6.47 m/s. What is the uncertainty in v¯x,0\bar{v}_{x,0}? ±0.01\pm 0.01 m/s±0.04\pm 0.04 m/s±0.06\pm 0.06 m/s±0.08\pm 0.08 m/s±0.09\pm 0.09 m/s±0.11\pm 0.11 m/s±0.15\pm 0.15 m/s±0.21\pm 0.21 m/sClear my selection

BlackTom AI · Accepted Answer

We’re given a dataset of vx,0 values: 6.34, 6.53, 6.43, 6.48, 6.58 m/s, with a reported mean v̄x,0 = 6.47 m/s. To estimate the uncertainty in the mean, we can use the standard error of the mean, which is s / sqrt(n), where s is the sample standard deviation and n is the number of measurements.

First, compute deviations from the mean: (6.34 − 6.47) = −0.13, (6.53 − 6.47) = 0.06, (6.43 − 6.47) = −0.04, (6.48 − 6.47) = 0.01, (6.58 − 6.47) = 0.11 m/s.

Next, square these deviations: 0.0169, 0.0036, 0.0016, 0.0001, 0.0121. Sum = 0.0343. This sum is used to compute the sample variance: s² = sum / (n − 1) = 0.0343 / 4 ≈ 0.008575. Therefore s ≈ sqrt(0.008575) ≈ 0.0926 m/s.

With n = 5 measurements, the standard error of the mean is SE = s / sqrt(n) ≈ 0.0926 / 2.236 ≈ 0.041 m/s, which rounds to ±0.04 m/s.

Now we evaluate each option:
- Option ± 0.01 m/s: This would imply a much smaller spread than observed; the individual deviations (up to 0.13 m/s) indicate the mean cannot be this tightly constrained. So this is too small.
- Option ± 0.04 m/s: This matches the calculated standard error of the mean using the sample standard deviation, ≈0.041 m/s, which is consistent with the data quality and number of measurements. This is a plausible uncertainty value.
- Option ± 0.06 m/s: Slightly larger than the calculated SE; while not impossible in different contexts, the direct calculation based on the data yields about 0.04 m/s, so 0.06 m/s would overestimate the precision.
- Option ± 0.08 m/s: This is roughly double the computed SE; it would significantly overstate the precision relative to the observed scatter for five measurements, making it unlikely.
- Option ± 0.09 m/s: Very close to the computed s but larger than SE; since the common practice for the uncertainty in a mean is SE rather than the full scatter s, this would be an overestimate for the mean’s uncertainty in typical reporting.
- Option ± 0.11 m/s: This is even larger and would correspond to using the full spread rather than the standard error, which is not appropriate for the uncertainty of the mean value itself in this context.
- Option ± 0.15 m/s: Clearly too large given the small sample and the calculated SE; this would imply far less confidence than warranted by the data.
- Option ± 0.21 m/s: The largest option, far exceeding the observed dispersion of the sample and not consistent with the standard error of the mean for five measurements.

In summary, the calculation based on the data leads to a standard error of the mean around 0.041 m/s, so the option that best matches this result is ± 0.04 m/s.

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