SS 024a – Exam #2 - Department of Statistical and Actuarial Sciences
... 4. An efficiency expert wishes to determine the average time that it takes to drill three holes in a certain metal clamp. How large a sample will he need to be 99% confident that his sample mean will be within 15 seconds of the true mean? Assume that it is known from previous drilling studies that t ...
... 4. An efficiency expert wishes to determine the average time that it takes to drill three holes in a certain metal clamp. How large a sample will he need to be 99% confident that his sample mean will be within 15 seconds of the true mean? Assume that it is known from previous drilling studies that t ...
One sample statistical tests, continued…
... The bigger the sample size (i.e., the bigger the sample size used to estimate ), then the closer t becomes to Z. If n>100, t approaches Z. ...
... The bigger the sample size (i.e., the bigger the sample size used to estimate ), then the closer t becomes to Z. If n>100, t approaches Z. ...
17. Inferential Statistics
... To compare survey data from Nominal or Ordinal Scales -without a Mean Score, so use a Nonparametric Tests. Chi Square tests the difference in Frequency Distributions of two or more groups. ...
... To compare survey data from Nominal or Ordinal Scales -without a Mean Score, so use a Nonparametric Tests. Chi Square tests the difference in Frequency Distributions of two or more groups. ...
6. Measures of central tendency and variation.
... Outliers are observations above Q3 + 1.5IQR or below Q1 − 1.5IQR. Also, serious outliers are observations above Q3 + 3IQR or below Q1 − 3IQR. In our example we do not have any outliers since Q3 + 1.5IQR = 60 + 1.5(32) = 108 and Q1 − 1.5IQR = 28 − 1.5(32) = −20. Now we can construct the box plot. ...
... Outliers are observations above Q3 + 1.5IQR or below Q1 − 1.5IQR. Also, serious outliers are observations above Q3 + 3IQR or below Q1 − 3IQR. In our example we do not have any outliers since Q3 + 1.5IQR = 60 + 1.5(32) = 108 and Q1 − 1.5IQR = 28 − 1.5(32) = −20. Now we can construct the box plot. ...
