*Continue to analyze your previous data?*Analysis Name Save Date and Time Row & Column Remove? New analysis *What's your objective?*Objective Example for Business Example for Researcher Paid? Want to predict data Want to predict our sales with the ads expenses on an state-by-state basis for the smartphone we sell State,Ads Expenses,Sales California,100,2100 Texas,130,2500 New York,150,2800 Florida,120,2300 Illinois,110,2300 Pennsylvania,140,2400 Sales is shown by the following formula. Sales = 11.129 * Ads Expenses + 971.429 Pretty good accuracy! The measured values (x marked) and the predicted values (solid line) are shown by the following graph. Number of measurement,Value of attribute A,Value of object 1st,100,2100 2nd,130,2500 3rd,150,2800 4th,120,2300 5th,110,2300 6th,140,2400 Value of object is shown by the following formula. Value of object = 11.129 * Value of attribute A + 971.429 Pretty good accuracy! The measured values (x marked) and the predicted values (solid line) are shown by the following graph. The partial regression coefficient are 11.428571428571429, 971.4285714285713, Contribution ratio is 0.8163265306122442. If a contribution rate is more than 0.8, we show 'pretty good' as the accuracy of the prediction. If it is more than 0.5 and less than 0.8, we show 'good'. If it is less than 0.5, you have to analyze data in another way. Want to know how much data I need to prepare Want to know the number of people I should pick up to get an accurate result from a questionaire targeting smartphone users in our sales area Yes Acceptable Error,Standard Deviation (a range of data to include about 68% of the total centering on a mean) 5,20 (You can skip the first line) Please analyze at least 62 numbers of the measured data. Acceptable Error,Standard Deviation (a range of data to include about 68% of the total centering on a mean) 5,20 (You can skip the first line) Please analyze at least 62 numbers of the measured data. This is a sample size to get a 95% confidence interval in case a population is assumed to follow a normal distribution, an error is 5 and standard deviation is 20. Want to estimate a mean Want to estimate hours of use for a smartphone from a questionaire targeting smartphone users in our sales area Responder,Hours of use Andy,1.5 Billy,2.0 Chris,3.0 Duolon,1.8 Eiji,2.2 Foxy,2.1 We can estimate the population of Sales Volume about Responder at the following range 1.5690191472127581 to 2.630980852787242 If a range is too big, it is possible for the sample size to be low or for event itself to have a great variability You can check a sample size at the above 'Want to know how much data I need to prepare'. Number of measurement,Measured value 1st,1.5 2nd,2.0 3rd,3.0 4th,1.8 5th,2.2 6th,2.1 We can estimate the population of Measured value about Number of measurement at the following range 1.5690191472127581 to 2.630980852787242 If a range is too big, it is possible for the sample size to be low or for event itself to have a great variability You can check a sample size at the above 'Want to know how much data I need to prepare'. This is a 95% confidence interval in case a population is assumed to follow a normal distribution and a population variance is unknown If the number of data (n) is more than 25 except for the label line, we use a normal distribution table. If the number of data (n) is less than 25, we use a t-distribution table with a degree of freedom n-1. Want to know how much data I need to prepare (estimation of ratio) Want to know the number of people I should pick up to get an accurate result from a questionaire targeting smartphone users in our sales area Reduced error rate (0-1.0) 0.05 (You can skip the first line) Please analyze at least 385 numbers of the measured data. Acceptable Error,Standard Deviation (a range of data to include about 68% of the total centering on a mean) 5,20 (You can skip the first line) Please analyze at least 62 numbers of the measured data. This is a sample size to get a 95% confidence interval in case a population is assumed to follow a normal distribution and an error is 0.05. Want to estimate a ratio Want to estimate a ratio of the number of smartphone users that answered that they are satisfied with your service from a questionaire in our sales area Yes Overall number,A ratio for estimation (0-1.0) 500,0.25 We can estimate the population at the following range 0.2120447632071673 to 0.2879552367928327 If a range is too big, it is possible for the sample size to be low or for event itself to have a great variability You can check a sample size at the above 'Want to know how much data I need to prepare (For estimation of ratio)' Overall number,A ratio for estimation (0-1.0) 500,0.25 We can estimate the population at the following range 0.2120447632071673 to 0.2879552367928327 If a range is too big, it is possible for the sample size to be low or for event itself to have a great variability You can check a sample size at the above 'Want to know how much data I need to prepare (For estimation of ratio)' This is a 95% confidence interval in case a population is assumed to follow a normal distribution and a population variance is unknown. Want to know if there is a gap between data and theoretical values Want to know if the models smartphone users have are weighted with specific ones from sales volumes in our sales area Yes Models,Sales volume,Theoretical values Model A,125,100 Model B,110,100 Model C,90,100 Model D,100,100 Model E,100,100 Model F,90,100 Model G,85,100 This follows a theory! The measured values (x marked) and the theoretical values (dots) are shown by the following graph. Research object,Observed Number,Expected Values A,125,100 B,110,100 C,90,100 D,100,100 E,100,100 F,90,100 G,85,100 This follows a theory! The measured values (x marked) and the theoretical values (dots) are shown by the following graph. A goodness of fit test was done by using a chi-square distribution The test statistic is 11.5. If n is the number of data except for the label line, we uses a chi-square distribution table with a degree of freedom n-1 and a significance level 0.05. Want to compare two groups Want to confirm that we have a advantage over a competitor in customer satisfaction for our smartphones from its survey Yes Responder,Our CS,Competitor CS Andy,91,89 Billy,97,95 Chris,91,92 Duolon,93,90 Eiji,98,96 Foxy,93,92 Obviously Our CS is a bigger mean. Our CS (x marked) and Competitor CS (dots) are shown by the following graph. Number of measurement,Measured Values (new method),Measured Values (old method) 1st,90,89 2nd,97,95 3rd,91,92 4th,93,90 5th,98,96 6th,93,92 Obviously Measured Values (new method) is a bigger mean. Measured Values (new method) (x marked) and Measured Values (old method) (dots) are shown by the following graph. We did a test of the difference of the paired population means under the null hypothesis that there is no difference between them. The test statistic is 2.6655699499159153. If the number of data (n) is more than 25 except for the label line, we use a normal distribution table. If the number of data (n) is less than 25, we use a t-distribution table with a degree of freedom n-1. Both significance levels are 0.05. Want to know if the data has a outlier Want to exclude a outlier that is possible to be measured wrongly in parts test for smartphones Responder,Hours of use Andy,1.5 Billy,4.9,unexpected Chris,2.0 Duolon,2.5 Eiji,3.0 Foxy,2.3 Billy,4.9 is a outlier. Exclude it from the data. Number of measurement,Measured Values 1st,1.5 2nd,4.9,unexpected 3rd,2.0 4th,2.5 5th,3.0 6th,2.3 2nd,4.9 is a outlier. Exclude it from the data. Assuming that a population follows a normal distribution, we did a Smirnov-Grubs test. The test statistic is 1.851421626658838. If n is the number of data except for the label line, we use a Smirnov-Grubs test table with a significance level 0.05. Want to take the black and white out of data Want to exclude inferior parts in a parts test of smartphones Yes Production Number,Score,Group 01,90,good 02,80,inferior 03,85,good 04,86,inferior 05,82,inferior 06,86,good 07,85,inferior The boundary value between good/inferior is 85.21765528644033. You can regard it as 'good' if you get a new bigger Score than this value. Object,Measured data,Group 01,90,group A 02,80,group B 03,85,group A 04,86,group B 05,82,group B 06,86,group A 07,85,group B The boundary value between group A/group B is 85.21765528644033. You can regard it as 'group A' if you get a new bigger Mewasured data than this value. The boundary is a value that univariate Mahalanobis' generalized distance is the same at each group *Enter data into a textarea**Save your analysis?*

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