Hey...I just want to make sure my answers to these questions are correct...
1. d=2.507 g/1.22 mL = 2.054918033 g mL^-1
explain how both the rules for significant figures and the random error calculation (p(d) = +-0.34 g mL^-1) indicate that the digits after the hundreths place have no meaning in this measurement.
Answer: The rules for significant figures and the random error calculation indicate the digits after the hundredths place have no meaning in this measurement because if the experiment was performed again, the same exact number would not be calculated again because of possible fluctuations and irreproducibility of an instrument.
2. although the beads used for experimental mass and volume determinations are to be chosen randomly, certain care should be taken not to choose beads that would introduce systematic errors. how would the accuracy of the volume determination be affected (answer \\\"high\\\", \\\"low\\\", or \\\"no significant change\\\") by using beads that are...
warped (nonspherical)
A: no sig change
hollow
A: high
slightly chipped
A: low
slightly chipped with a large air bubble that has become attached to the outside of the bead
A: high
3. assume that the estimates of the standard deviation for the mean mass of a series of three measurements, a serious of six measurements, and a series of nine measurements were identical. why will the confidence intervals for three mean masses differ? which different is greater: that between the CI for three measurements and the CI for six measurements; or that between the CI for six measurements and the CI for nine measurements? Briefly explain.
A: The confidence intervals for the three mean masses will differ because as the number of beads increases, the confidence interval will decrease. There is a greater possibility of getting a better confidence interval.
The difference between the CI and three measurements is the greatest because there is little degree of freedom for the selection.
The difference between the CI for six measurements and the CI for nine measurements is smaller because the degree of freedom for the selection is greater.
1. Your answer is partly correct. The rules for significant figures indicate that the digits after the hundredths place have no meaning because they are not reliably measured or significant. However, the random error calculation also suggests that these digits have no meaning because the value of the random error (±0.34 g mL^-1) exceeds the value of the digits after the hundredths place (0.054918033 g mL^-1). So, both the rules for significant figures and the random error calculation support the conclusion that the digits after the hundredths place have no meaning in this measurement.
2. Your answers are incorrect. Using warped (nonspherical) beads would introduce systematic errors and affect the accuracy of the volume determination, causing it to be either high or low. Using hollow beads would result in a higher volume determination, so the answer is "high". Using slightly chipped beads would introduce errors but to a lesser extent, resulting in a slightly lower volume determination, so the answer is "low". Lastly, using slightly chipped beads with a large air bubble attached to the outside would greatly affect the volume determination, causing it to be higher than normal, so the answer is "high".
3. Your answer is incorrect. The confidence intervals for the mean masses will differ because the sample size affects the precision of the estimate. As the sample size increases, the confidence interval becomes smaller, indicating a greater precision. Therefore, the difference between the confidence interval for six measurements and the confidence interval for nine measurements is greater than the difference between the confidence interval for three measurements and the confidence interval for six measurements. The larger the sample size, the more accurate the estimation of the mean.
1. Your answer is partially correct. The rules for significant figures indicate that the digits after the hundredths place have no meaning because the given value has one decimal place (1.22 mL) and therefore the answer should also have one decimal place (2.0 g/mL). The random error calculation (p(d) = ±0.34 g mL^-1) indicates that the measurement can vary by up to 0.34 g mL^-1, which is significantly larger than the calculated value of 2.054918033 g mL^-1. This means that the digits after the hundredths place are not reliable and do not contribute to the accuracy of the measurement.
2. Your answers are correct.
- Warped (nonspherical) beads: No significant change in the accuracy of the volume determination. The measurement of volume is not affected by the shape of the beads as long as their mass and diameter are accurately measured.
- Hollow beads: The accuracy of the volume determination would be high. Hollow beads have a higher volume-to-mass ratio compared to solid beads of the same size. This means that for a given mass, the volume would be larger for hollow beads, leading to a higher calculated volume.
- Slightly chipped beads: The accuracy of the volume determination would be low. Chipped beads may result in inaccurate measurements of both mass and diameter, leading to incorrect volume calculations.
- Slightly chipped beads with a large air bubble attached: The accuracy of the volume determination would be high. The presence of a large air bubble would increase the volume without significantly affecting the mass, resulting in an overestimation of the volume.
3. Your answer is partially correct. The confidence intervals for mean masses will differ because the confidence interval is inversely proportional to the sample size. As the sample size increases, the confidence interval becomes narrower, indicating a more precise estimate of the true mean. Therefore, the CI for three measurements will be wider than the CI for six measurements, and the CI for six measurements will be wider than the CI for nine measurements. The difference between the CI for three measurements and the CI for six measurements will be greater because the increase in sample size from three to six provides a larger reduction in uncertainty compared to the increase in sample size from six to nine.
1. The rules for significant figures indicate that only the digits that are known with certainty should be included in the measurement. In this case, the original value of d has three significant figures (2.507), and the measurement of d as calculated is rounded to four significant figures (2.055). Therefore, the digits after the hundredths place (8 and 3) have no meaning since they are not known with certainty.
The random error calculation (p(d) = ±0.34 g mL⁻¹) indicates the range of possible values for the true value of d. The random error implies that the actual value of d could be anywhere within the range of ±0.34 g mL⁻¹ of the calculated value. Since the random error is ±0.34 g mL⁻¹, which is larger than the difference between the calculated value and the digits after the hundredths place (0.055 g mL⁻¹), it further supports the idea that those digits have no meaning in this measurement.
2. If beads that are warped (nonspherical) are used for volume determinations, the accuracy of the volume determination would not be significantly affected. Since the beads are chosen randomly, the variations in the shape of the beads would average out over multiple measurements. Therefore, there would be no significant change in accuracy.
If hollow beads are used, the accuracy of the volume determination would be high. Hollow beads have additional empty space inside, which would result in an underestimation of the actual volume occupied by the beads.
If slightly chipped beads are used, the accuracy of the volume determination would be low. The presence of chips or irregularities can cause the beads to have a larger volume than they should, resulting in an overestimation of the actual volume.
If slightly chipped beads with a large air bubble attached to the outside are used, the accuracy of the volume determination would be high. The presence of the large air bubble would cause the apparent volume of the beads to be significantly increased, resulting in an overestimation of the actual volume.
3. The confidence intervals for the mean masses will differ because the sample size affects the precision of the mean estimate. As the number of measurements increases, the standard deviation of the mean decreases, leading to a narrower confidence interval. This is because a larger sample size provides more information and reduces the uncertainty in the measurement.
In this case, since the estimates of the standard deviation for the three series of measurements are identical, the confidence intervals for three, six, and nine measurements will differ based on the formula for the confidence interval calculation. The difference between the CI for three measurements and the CI for six measurements will be greater than the difference between the CI for six measurements and the CI for nine measurements because adding more measurements decreases the standard deviation of the mean and hence narrows the confidence interval. The increase in precision from three measurements to six measurements will be greater than the increase from six measurements to nine measurements.