Descriptive Statistics: Income ($1000), Size, Credit Balance ($) Variable Mean StDev Minimum Median Maximum Range Mode Income ($1000) 43.74 14.64 21.00 43.00 67.00 46.00 55 Size 3.420 1.739 1.000 3.000 7.000 6.000 2 Credit Balance($) 3976 932 1864 4090 5678 3814 3890 Variable Mode Skewness Kurtosis Income ($1000) 4 0.05 -1.29 Size 15 0.53 -0.72 Credit Balance($) 2 -0.14 -0.74 The statistics indicates that the mean income is 43.74; the median income is 43.00 and the mode income is 55. The standard deviation is given approximately as 1.74. Maximum income is 67,000 and the minimum income is 21,000 and a Standard Deviation of 14.64. The second individual variable is household size the mean household size of the sample is 3.42, the median household size of the sample is 3 and the mode household size of the sample is 2 and the standard deviation is 1.739. The maximum household size is 7 and the minimum household size is
What is the mean of the above values? To calculate the mean add all of the values in the data together and divide by the number of values. 2.0+3.7+4.9+5.0+5.7+6.7+8.5+9.0=45.5/8=5.6875= 5.7 is the mean of the numbers above. What is the mode of the above values? The mode is the value that occurs most often.
c. Discuss the factors which might cause variation in the weight of Tootsie Rolls during manufacture. Sample mean: 3.3048 Standard deviation: .1320 90 percent confidence interval (taken from text): 1.812 a. X ± z σ√n 3.3048 ± .1320√10 = .0364 3.3048-.0364 = 3.2684 3.2684 to 3.3412 3.3048+.0364 = 3.3412 b. n =( zσE) ^ 2 (1.1812*.1320) / .03 ^2 = 63.57 round to 64
If markets for Apogee's goods and services exist in multiple locations, having one office hardly makes sense. Multiple offices should be opened to increase profit. The author could do a couple of things to make this argument stronger. For example, there is no evidence that shows the profitability of all the field offices is declining. There is also no evidence that shows the marketability of the field offices.
HW assignment #1 Classify each of the following types of data for questions 1-3: 1) Ethnicity is a Qualitative variable (Nominative) 2) Length is a Quantitative variable. 3) Temperature (Celsius) is an Interval Variable 4) Make a Stem and Leaf plot from the following data 45, 50, 51, 110, 92, 71, 105, 56, 51, 62, 64, 77, 65, 55, 55, 59, 48, 73, 99, 116, 56 4 5 8 5 0 1 1 5 5 6 6 9 6 2 4 5 7 1 3 7 9 2 9 10 5 11 0 6 5) A manager of a car rental company took a random sample of 100 business days over the last fiscal year and recorded the number of cars rented per day. The frequency distribution for the data is given below. Interval Frequency RF CF CRF [20, 25) 4 .04 4 4/100 = .04 [25, 30) 11 .11 15 15/100 = .15 [30, 35) 23 .23 38 38/100 = .38 [35, 40) 31 .31 69 69/100 = .69 [40, 45) 15 .15 84 84/100 = .84 [45, 50) 10 .10 94 94/100 = .94 [50, 55) 6 .06 100 100/100 = 1 100 1 a) Calculate the Relative Frequency, Cumulative Frequency, and the Cumulative Relative Frequency. b) Draw a histogram and frequency polygon plot, and a relative frequency pie chart.
Yes, it is significant because as indicated by the asterisk, p <0.05 is the least acceptable value for statistical significance. 3. Which t ratio listed in Table 3 represents the smallest relative difference between the pretest and 3 months? Is this t ratio statistically significant? What does this result mean?
To find the mean of the sample data set the formula must be used: Sample mean x̄ = (∑x)/N = 446.1/30= 14.87 According to Larson and Farber, 2009, the median of a data set is the value that lie in the middle of the data when the data set is ordered. The given numbers in the table are in order, there are thirty entries (an even number). The median is the mean of the two middle entries. Median = (14.8+14.8)/2 = 14.8 According to Larson and Farber, 2009, the standard deviation is the difference between the entry data and the entry mean. Sample standard deviation x=√(∑(x-µ)^2 )/(N-1) = √8.783/(30-1) = √0.302862 = 0.5503 Construct a 95% Confidence Interval for the ounces in the bottles.
Given: sides, 5,6,7 in miles Island A = 78.46° Island B = 135.58° Island C = 57.12° Given the information… From Island B I would travel a Northwest Bearing to Island C. 8. Heron’s formula: sq.root of {s(s-a)(s-b)(s-c)} Sqrt((a^2+b^2+c^2)^2 - 2 (a^4+b^4+c^4))/4 = sqrt ((25+25+16)^2-2(625+625+256))/4 = sqrt ((66)^2-3012)/4 = sqrt (4356-3012)/4 = 36.66060556/4 = 9.16515139 square feet 9 feet
Assignment: M&M Project Part 3 M&M’s Project Part 3 Strayer University Math 300 Statistics Professor Ahmed Rawish March 14, 2012 Construct a 95% Confidence Interval for the proportion of blue M&M’s candies. 95% Confidence Interval for proportion is given by [pic] where p = x/n = 847/4179 = 0.202680067, [pic]= 1.959963985, n = 4179 Therefore, CI is given by, [pic] = (0.190492031, 0.214868103) Thus with 95% confidence we can claim that the proportion of blue M&Ms® candies is within (0.190492031, 0.214868103). Details |Confidence Interval Estimate for Proportion | | | | |Data | | |Sample Size |4179 | |Number of Successes |847 | |Confidence Level |95% | | | | |Intermediate Calculations | |Sample Proportion |0.202680067 | |Z Value |-1.959963985 | |Standard Error of the Proportion |0.0062185 | |Interval Half Width |0.012188036 | | | | |Confidence Interval | |Interval Lower Limit |0.190492031 | |Interval Upper Limit |0.214868103 | 3 pts. Construct a 95% Confidence Interval for the proportion of orange M&Ms® candies. 95% Confidence Interval for proportion is given by [pic] where p = x/n =
Current cost report isn’t useful for evaluation of formed vs purchased bales prices. Neither supplementary nor summary cost report contains information about wages, employees and allocation of indirect labor. The second most costly item in Depot Oh, named other OH, isn’t expanded by reducing the ability to analyze additional data in report and come with a fair conclusion on the state of the company cost accounting. Allocated division overhead is also not expanded, however the amount reduced by 10% which could be a question of efficiency of plants. 4.