The normality of NaOH obtained was 0.0172, 0.0187, 0.0185, 0.189, and 0.0183 N. The theoretical normality of NaOH was 0.02 N. Please calculate:
1) Standard deviation based on the true mean:
2) Standard deviation based on the experimental mean: SD
3) Coefficient of variation (CV)
4) Confidence Interval of 95% certainty
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1) Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.
Standard deviation = square root of variance
2) Find the mean first = sum of scores/number of scores
Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.
Standard deviation = square root of variance
3) https://www.google.com/search?client=safari&rls=en&q=Coefficient+of+variation&ie=UTF-8&oe=UTF-8
4) 95% = mean ± 2 SD
I'll let you do the calculations.
We balance our postings with other factors in our lives.
To calculate the values you need, follow these steps:
1) Standard deviation based on the true mean:
a) Calculate the mean of the measured normalities: Add up all the normality values and divide by the number of measurements.
Mean (x̄) = (0.0172 + 0.0187 + 0.0185 + 0.0189 + 0.0183) / 5
b) Calculate the sum of the squared differences between each measurement and the mean:
(0.0172 - x̄)^2 + (0.0187 - x̄)^2 + (0.0185 - x̄)^2 + (0.0189 - x̄)^2 + (0.0183 - x̄)^2
c) Divide the sum of the squared differences by the number of measurements:
Variance (σ²) = sum of squared differences / number of measurements
d) Calculate the square root of the variance to get the standard deviation (σ):
Standard Deviation (σ) = √(variance)
2) Standard deviation based on the experimental mean:
This is the same as the previous calculation because both the true mean and experimental mean are the same in this case.
3) Coefficient of variation (CV):
The coefficient of variation can be calculated by dividing the standard deviation by the mean and multiplying by 100.
Coefficient of Variation (CV) = (σ / x̄) * 100
4) Confidence Interval of 95% certainty:
a) Calculate the standard error, which is the standard deviation divided by the square root of the number of measurements:
Standard Error (SE) = σ / √(number of measurements)
b) Calculate the margin of error by multiplying the standard error by the appropriate t-value from a t-distribution table for a 95% confidence level with (n-1) degrees of freedom.
c) Subtract and add the margin of error to the mean to get the confidence interval:
Confidence Interval = Mean ± Margin of Error
Let's calculate these values using the given data:
1) Standard deviation based on the true mean:
Mean (x̄) = (0.0172 + 0.0187 + 0.0185 + 0.0189 + 0.0183) / 5 = 0.0183 N
Variance (σ²) = [(0.0172 - 0.0183)^2 + (0.0187 - 0.0183)^2 + (0.0185 - 0.0183)^2 + (0.0189 - 0.0183)^2 + (0.0183 - 0.0183)^2] / 5
= [0.000016 + 0.000004 + 0.000004 + 0.000036 + 0] / 5
= 0.000012 N²
Standard Deviation (σ) = √(variance) = √(0.000012) ≈ 0.00346 N
2) Standard deviation based on the experimental mean:
Since the experimental mean is the same as the true mean, the standard deviation remains the same: 0.00346 N.
3) Coefficient of variation (CV):
Coefficient of Variation (CV) = (σ / x̄) * 100 = (0.00346 / 0.0183) * 100 ≈ 18.9%
4) Confidence Interval of 95% certainty:
Standard Error (SE) = σ / √(number of measurements) = 0.00346 / √(5) ≈ 0.001546 N
T-value for 95% confidence level, with (n-1) degrees of freedom = 2.776 (approximated from the t-distribution table)
Margin of Error = Standard Error * T-value = 0.001546 * 2.776 ≈ 0.00429 N
Confidence Interval = Mean ± Margin of Error = 0.0183 ± 0.00429 N
Therefore, the calculated values are:
1) Standard deviation based on the true mean: 0.00346 N
2) Standard deviation based on the experimental mean: 0.00346 N
3) Coefficient of variation (CV): 18.9%
4) Confidence Interval of 95% certainty: 0.01301 N to 0.02359 N