1 / 23

Are our results reliable enough to support a conclusion?

Are our results reliable enough to support a conclusion?. 810. 811. Imagine we chose two children at random from two class rooms…. … and compare their height …. 810. 811. … we find that one pupil is taller than the other. WHY?. 810. 811.

Download Presentation

Are our results reliable enough to support a conclusion?

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Are our results reliable enough to support a conclusion?

  2. 810 811 Imagine we chose two children at random from two class rooms… … and compare their height …

  3. 810 811 … we find that one pupil is taller than the other WHY?

  4. 810 811 There is a significant difference between the two groups, so pupils in 811 are taller than pupils in 810 REASON 1: YEAR 7 YEAR 11

  5. 810 811 TITCH HAGRID (Year 9) (Year 9) By chance, we picked a short pupil from 810 and a tall one from 811 REASON 2:

  6. How do we decide which reason is most likely? MEASURE MORE STUDENTS!!!

  7. 810 811 If there is a significant difference between the two groups… … the average or mean height of the two groups should be very… … DIFFERENT

  8. 810 811 If there is no significant difference between the two groups… … the average or mean height of the two groups should be very… … SIMILAR

  9. Living things normally show a lot of variation, so… Remember:

  10. 810 Sample 811 Sample Average height = 168 cm Average height = 162 cm It is VERY unlikely that the mean height of our two samples will be exactly the same Is the difference in average height of the samples large enough to be significant?

  11. 16 16 810 Sample 14 14 12 12 10 10 Frequency Frequency 8 8 6 6 4 4 2 2 180-189 160-169 140-149 140-149 160-169 170-179 150-159 170-179 150-159 Height (cm) 811 Sample 180-189 Height (cm) We can analyse the spread of the heights of the students in the samples by drawing histograms Here, the ranges of the two samples have a small overlap, so… … the difference between the means of the two samples IS probably significant.

  12. 16 16 810 Sample 14 14 12 12 10 10 Frequency Frequency 8 8 6 6 4 4 2 2 180-189 160-169 140-149 140-149 160-169 170-179 150-159 170-179 150-159 Height (cm) 811 Sample 180-189 Height (cm) Here, the ranges of the two samples have a large overlap, so… … the difference between the two samples may NOT be significant. The difference in means is possibly due to random sampling error

  13. You can calculate standard deviation using the formula: (Σx)2 Where: Sx is the standard deviation of sample Σ stands for ‘sum of’ x stands for the individual measurements in the sample n is the number of individuals in the sample Σx2 - Sx = n n - 1 To decide if there is a significant difference between two samples we must compare the mean height for each sample… … and the spread of heights in each sample. Statisticians calculate the standard deviation of a sample as a measure of the spread of a sample

  14. ( x1– x2 ) Where: x1 is the mean of sample 1 s1 is the standard deviation of sample 1 n1 is the number of individuals in sample 1 x2 is the mean of sample 2 s2 is the standard deviation of sample 2 n2 is the number of individuals in sample 2 t = (s1)2 (s2)2 + n1 n2 Student’s t-test The Student’s t-test compares the averages and standard deviations of two samples to see if there is a significant difference between them. We start by calculating a number, t t can be calculated using the equation:

  15. 810: x1 = 811: x2 = x2 – x1 = 168.27 – 161.60 = Worked Example: Random samples were taken of pupils in 810 and 811 Their recorded heights are shown below… Step 1: Work out the mean height for each sample 161.60 168.27 Step 2: Work out the difference in means 6.67

  16. 811: 810: (s2)2 (s1)2 = = n2 n1 Step 3: Work out the standard deviation for each sample 810: s1 = 10.86 811: s2 = 11.74 Step 4: Calculate s2/n for each sample 10.862÷ 15 = 7.86 11.742÷ 15 = 9.19

  17. Step 5: Calculate (s2)2 (s2)2 (s1)2 (s1)2 (s1)2 = = (7.86 + 9.19) = + + + n2 n2 n1 n1 n1 (s2)2 n2 t = 6.67 = 4.13 x2 – x1 4.13 Step 6: Calculate t (Step 2 divided by Step 5) 1.62

  18. Step 7: Work out the number of degrees of freedom d.f. = n1 + n2 – 2 = 15 + 15 – 2 = 28 Step 8: Find the critical value of t for the relevant number of degrees of freedom Use the 95% (p=0.05) confidence limit Critical value = 2.05 Our calculated value of t is below the critical value of 2.05 (df = 28 and alpha level of significance = 0.05); therefore, there is no statistically significant difference between the height of students in samples from 810 and 811.

  19. Do not worry if you do not understand how or why the test works Follow the instructions CAREFULLY You will NOT need to remember how to do this for your exam

  20. T test summary..thanks Mrs. Cole! • If your t value is less than the critical t value, then your data is NOT statistically significant. • If your t value is greater than the critical t value, then  your data IS statistically significant. • What statistically significant means (loosely) is that it is more likely than not that the results you’ve obtained from your sample data is NOT due purely to chance. • When you state your critical t value, it would also be a good idea to state the degrees of freedom and the significance level.  Just for example: Our calculated value of t is below the critical value of 2.05 (df = 28 and alpha level of significance = 0.05); therefore, there is no statistically significant difference between the height of students in samples from 810 and 811.

  21. “Correlation does not imply causation" is a phrase used in science and statistics to emphasize that correlation between two variables does not automatically imply that one causes the other

  22. Correlation does not imply causation

  23. To determine cause requires experimental evidence!

More Related