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Published bySheila Warren Modified over 9 years ago
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Confidence Intervals and Significance Testing in the World of T Welcome to the Real World… The World of T T
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T is used for testing means (averages)… Looking for Statistically Significant differences Aquaculture Calculating and comparing Tank Flow Rate to Ideal Flow Rate HMMM… T – Testing in Research????
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T is a density curve Symmetric about Zero, single peaked, “bell” shaped T’s variation depends on sample size Remember, samples become less variable as they get larger Degrees of Freedom T makes an adjustment for each sample size by changing the degrees of freedom Basically gives us a new T to work with for each sample size!! Let me break this down really simply for you… For T- Testing, we’re still testing for Population Means, but we only need SAMPLE data!! Let’s See What Makes T, T… Check Me Out!! Wow! A tailor made T for each sample!!
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Standard Error Sample Standard Dev. With n-1 Degrees of Freedom One Sample T Statistic This is that personal touch for each different sample…
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P-Value Area to the right of t Area to the left of –t 2(P) for two-sided This is for the T- Table… Let’s Practice Degrees of Freedom (df) Left hand column of chart Different T-Distribution for each sample size Larger the sample, the closer to Normal the T distribution T-Statistic Located in MIDDLE of chart Leads to the p-value or vice versa
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Find the t-statistic for the following: 1) 5 dof; p =.05 (right) 2) n = 22; p =.99 (left) 3) 80% CI; n = 18 Find the p-value for the following: 1) 5 dof; t = 3.365 2) n = 12; t = 1.856 3) n = 67; t = 2.056 t = 2.015 t = 2.518 t = 1.333 p =.01.025 < p <.05.02 < p <.025 Notice the t-statistic is limited to certain values on your table!!! What happens if you get a T that’s not on your table? Then What? You will simply say you’re p-value is BETWEEN 2 values!!
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With these tests you are given an alpha level against which you test your p-value *(Standard level =.05): p ≤ – Reject the null; accept the H a p > – Fail to reject the null H a : µ > µ 0 H a : µ < µ 0 H a : µ ≠ µ 0
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You need to show this check of conditions after you write your hypotheses… Graph your distribution for samples less than 40 to determine level of normality!!! If your HISTOGRAM is skewed, either scrap the t-test or talk about the questionability of the results!! Sample SizeDistributionProof n <15Needs to be normal Histogram or Stem Plot 15 < n < 40No STRONG outliers or skewness Histogram or Stem Plot n < 40No restrictionsNot Needed
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For T – Tests (testing for population mean with sample mean and standard deviation… We use the Same Basic steps as in all Hypothesis Testing State the Ho and Ha in symbols and context Find the T-Statistic Find the p-value from the t-statistic w/ n-1 degrees of freedom Compare your p-value to the specified , and make your decision in context
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Mrs. Luniewski has claimed the ideal flow rate for fish growth is 22 L/min of water flow. You’ve decided to check to see if there’s a statistically significant difference between the ideal flow rate and the flow rate of your tank. You take a SRS of 50 rates from the past week and find your tank has an average flow rate of 14 with a standard deviation of 1.36. At a 5% significance level, is your flow rate significantly less than the ideal flow rate? Ho: µ = 22 L/min Ha: µ < 22L/m t = -2.4072 ( n – 1) df = 49 (round down to 40 for table) P is less than.0005 Since p is less than.0005, which is less than.05, we have statistically significant evidence that our flow rate is SIGNIFICANTLY less than the ideal rate. This would help us identify potential problems with the tank…
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Matched Pairs Test Used when taking same measurements on same media over different time period “Difference” between values is THE data H o = µ diff = 0 [µ diff = (µ 1 - µ 2 )] H a = µ diff or ≠ Flow Rate of Tank Pre- treatment Flow Rate of Tank Post-treatment Difference (Post – Pre) 129.52.5 1183 954 You use this difference column to get your Sample Mean and Sample Standard Deviation…
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Now you have applied a water treatment to the tanks, hoping to make a difference in the average flow rate of your tank. To check to see if the difference is statistically significant, you collect 3 measurements (not really enough!!) from the tanks at the exact same times they were collected pre-treatment. Test to see if there is a significant difference post-treatment. Because the measurements were taken on the same tank, at the same time, this would be considered a matched pairs test. You would need to adjust your hypotheses accordingly and use the difference between the data as your data source. Ho: µ diff = 0 Ha: µ diff < 0 (Increase in flow rate)
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Flow Rate of Tank Pre- treatment Flow Rate of Tank Post-treatment Difference (Post – Pre) 129.52.5 1183 954 Sample Average Difference = 3.17 Sample Std Dev =.7638 DF = 2. 005 < p <.01 REJECT Ho and conclude the treatment made a significant increase in flow rate…
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