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- Determine the Average Mass and the Standard Deviation for the following data: Mass of pennies in grams = 1.5021, 1.4736, 1.4966, 1.5164, 1.5420, 1.4699 g. Average Mass grams +/- Standard Deviation gramsThe following are the results for the % Cu of ore sample: 24.52%, 24.75%, 24.82%, 24.69%, 24.78% Calculate the mean. (A) 24.71% B) 0.12% 24.75% 0.47%A standard deviation of 0.044 is a good thing or a bad thing?
- Measurements of boiling temperatures of samples of a liquid (°C) are given below: 78.4 79.0 78.8 79.2 78.6 Calculate the meanCalculate the modeCalculate the standard deviationCalculate the standard deviation of the four trials below. Trial 1: 49.59 grams Trial 2: 51.25 grams Trial 3: 49.78 grams Trial 4: 50.19 grams9. The following are three sets of data for the atomic mass on antimony (Sb) from the work of Willard and McAlpine. Set 1 Set 2 Set 3 121.771 121.784 121.752 121.787 121.758 121.784 121.803 121.765 121.765 121.781 121.794 a) Determine the mean and standard deviation of each data set. b) Determine the 95% confidence intervals for each data set. c) Determine whether the 121.803 value in Set 1 is an outlier for that set. d) Pool the data and determine the overall mean and the pooled standard deviation. e) Compare the overall mean of the 11 data points with the currently accepted value. Report the absolute error and the percent relative error, assuming the currently accepted value is the true value.
- 9) The following results were obtained in the determination of Vitamin E mass in tablet samples. Sample 1 2 3 4 5 6 Weight of Vitamin E (mg) 249.1 249.9 251.2 250.2 244.8 251.0 i. Determine whether any data should be rejected or not at 90 % confidence level. ii. Calculate the mean and standard deviation of the data set above. ii. Express the mean weight of vitamin C at 99 % confidence interval.Calculate the standard deviation of the following results: 20,19,18,10,15,14An analysis of a sample containing mercury was performedusing a standard procedure and a new procedure. The samplewas analyzed 6 times by the new procedure and 8 times by thestandard procedure resulting in standard deviations of 0.355 and0.524 for the new and standard procedures, respectively. If thepooled standard deviation for the data was 0.66 and thetabulated F value at the 95% confidence level was 4.88, Thefollowing statement(s) are true:Select one or more:A. The calculated F value is 2.179B. We are confident that there is no significant statisticalC. difference between the results of the two methodsD. There is no significant statistical difference between theprecision of the two methodsThe calculated F value is 3.216
- STATISTICAL DATA IN CHEMICAL ANALYSIS: Calculate the mean and standard deviation of the data below. Write the proper reported data. 0.624, 0.613, 0.596, 0.607, 0.5828) Two different analytical methods are compared for determining Ca. The following are two sets of data. Set 1 Set 2 155.779 155.784 155.787 155.787 155.813 155.765 155.781 155.793 i. i. Determine the mean and the standard deviation in Set 1. Calculate the 95% confidence limit for data in Set 1. Identify a possible outlier in Set 2. Use the Q-test to determine whether it can be retained or rejected at 95% confidence level. ii.Calculate the standard deviation for an element whose % in a sample were found to be20.8, 21.6, 22.1, 22.0, 23.3, 21.9 and 22.8%
