Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Successfully reported this slideshow.

Like this presentation? Why not share!

- Lecture 7 Hypothesis Testing Two Sa... by Ahmadullah 29671 views
- Hypothesis testing; z test, t-test.... by Shakehand with Life 64361 views
- Principles of Statistical Inference... by Australian Bioinf... 991 views
- Seminar activity sheet social strat... by fatima d 264 views
- Un covenant civil political rights by fatima d 317 views
- Un covenant economioc social cultural by fatima d 272 views

No Downloads

Total views

1,276

On SlideShare

0

From Embeds

0

Number of Embeds

1

Shares

0

Downloads

19

Comments

0

Likes

1

No embeds

No notes for slide

- 1. Lecture 11 - The T-test C2 Foundation Mathematics (Standard Track) Dr Linda Stringer Dr Simon Craik l.stringer@uea.ac.uk s.craik@uea.ac.uk INTO City/UEA London
- 2. Hypothesis testing We use hypothesis testing to compare the mean of a very large data set, a population mean, with the mean of a sample data set, a sample mean. Z-test question: A lightbulb company says their lightbulbs last a mean time of 1000 hours with a standard deviation of 50. We think their lightbulbs last longer than this and propose a test at a 5% level of signiﬁcance. We buy 75 lightbulbs and they last a mean time of 1022 hours. The population mean is 1000 hours (A = 1000). The sample is the 75 light bulbs that we test (n = 75). The sample mean is 1022 hours (¯x = 1022.)
- 3. Z-test example A lightbulb company says their lightbulbs last a mean time of 1000 hours with a standard deviation of 50. We think their lightbulbs last longer than this and propose a test at a 5% level of signiﬁcance. We buy 75 lightbulbs and they last a mean time of 1022 hours. Hypotheses: H0 : µ = 1000, H1 : µ > 1000 Critical value: +1.65 Test statistic: ¯x−A σ/ √ n = 1022−1000 50/ √ 75 = 3.81 to 2 d.p. Decision: 3.81 > 1.65 so reject H0 Conclusion: The sample provides sufﬁcient evidence at 5% signiﬁcance level to reject null hypothesis; the lightbulbs last longer than 1000 hours.
- 4. Z-test summary You will be given 1. Population mean, A 2. Population standard deviation, σ 3. Signiﬁcance level (1% or 5%) 4. Sample mean, ¯x 5. Sample size, n 6. Quantifying word. You have to work out 1. Null hypothesis, alternative hypotheis 2. Critical value(s) 3. Test statistic 4. Decision - accept/reject H0 (sketch a picture if it helps) 5. Conclusion
- 5. The difference between a Z-test and a T-test In a Z-test the sample is large (n ≥ 25). You are given the sample mean and population or sample standard deviation In a T-test the sample is small (n < 25). You usually have to work out the sample mean and the sample standard deviation. Also in a T-test you have to work out the degrees of freedom to use in the critical value table. d.o.f. = n − 1
- 6. T-test summary You will be given 1. Population mean, A 2. Signiﬁcance level 3. Sample data set 4. Quantifying word. You have to work out 1. Null hypothesis (H0 : µ = A) and alternative hypotheis 2. Degrees of freedom, d.o.f. = n − 1 3. Critical value(s), look this up in the table 4. Sample mean, ¯x = Σx n 5. Sample standard deviation, s = x2−n¯x2 n−1 (MAKE SURE YOU CALCULATE s, not σ) 6. Test statistic, ¯x−A s/ √ n 7. Decision, accept/reject H0 (sketch a picture if it helps) 8. Conclusion, write this in words
- 7. The null hypothesis and the alternative hypothesis for the Z-test and T-test The null hypothesis is initially assumed to be true. It is H0 : µ = A where µ is ’population mean’ and A is the hypothetical value of the population mean The alternative hypothesis is either H1 : µ = A or H1 : µ < A or H1 : µ > A Sample data is collected and tested to see if it is consistent with the null hypothesis. If the sample mean is signiﬁcantly different from the population mean, H0 is rejected in favour of the alternative hypothesis, H1.
- 8. Signiﬁcance level The null hypothesis will always be tested to a given level of signiﬁcance. A 5% level of signiﬁcance means we are testing to see if the probability of getting the sample mean is less than 0.05. If the probability is less we reject the null hypothesis in favour of the alternative hypothesis. A 1% level of signiﬁcance translates to a probability of 0.01.
- 9. Critical value (T-test) The critical value is the boundary (or boundaries) of the rejection region(s). In a T-test this depends on the alternative hypothesis, signiﬁcance level and degrees of freedom (d.o.f. = n − 1, where n is the number of values in the data set) 5% signiﬁcance level 1% signiﬁcance level d.o.f. One-tailed Two-tailed One-tailed Two-tailed 2 2.92 4.30 6.97 9.93 3 2.35 3.18 4.54 5.84 4 2.13 2.78 3.75 4.60 5 2.02 2.57 3.37 4.03 6 1.94 2.45 3.14 3.71 7 1.90 2.37 3.00 3.50 8 1.86 2.31 2.90 3.36 9 1.83 2.26 2.82 3.25 10 1.81 2.23 2.76 3.17
- 10. Degrees of freedom (T-test) The degrees of freedom of a set of data is a way of measuring how the tests effect each other. If the data has size n and each sample does not effect any others the degree of freedom is n − 1. (This is usually the case with our data). Consider a bag containing 10 stones. If as a sample we pick out 10 stones our degree of freedom is 0 because the choice of the ﬁrst one constrains the possibilities for all others and the ﬁnal one is left with no choices. If as a sample we pick out 7 stones our degree of freedom is 3 because if we take out three stones before we start the choice of stones is unique. If as a sample we just pick out 1 stone our degree of freedom is 9.
- 11. H1 : µ = A If our alternative hypothesis is H1 : µ = A we are doing a two-tailed test and we have 2 critical values, one negative and one positive. The critical value is the boundary of the rejection region. For a 5% level of signiﬁcance we have the following picture: −2.31 2.31 The rejection (shaded) regions have a combined area of 0.05.
- 12. H1 : µ > A If our alternative hypothesis is H1 : µ > A we are doing a one-tailed test and we have 1 critical value which is positive. The critical value is the boundary of the rejection region. For a 5% level of signiﬁcance we have the following picture: 1.86 The rejection region has an area of 0.05.
- 13. H1 : µ < A If our alternative hypothesis is H1 : µ < A we are doing a one-tailed test and we have 1 critical value which is negative. The critical value is the boundary of the rejection region. For a 5% level of signiﬁcance we have the following picture: −1.86 The rejection region has an area of 0.05.
- 14. Sample mean and sample standard deviation The sample mean, ¯x is the mean ¯x = Σx n The sample standard deviation, s of a set of data is slightly different from the standard deviation σ. It is important to use the correct formula. For a T-test question ALWAYS use the sample variance formula s2 = x2 − n¯x2 n − 1 For a T-test question DO NOT USE the variance formula from Lecture 10 σ2 = x − ¯x n = x2 n − ¯x2 (We are using the sample data to work out an approximation to the population variance)
- 15. Test statistic and conclusion The test statistic is difference between the sample mean, ¯x and the (hypothetical) population mean A, divided by the standard error. The standard error is σ/ √ n for the Z-test and s/ √ n for the T-test, where n is the sample size, σ is the population standard deviation and s is the sample standard deviation. The T-test statistic is ¯x − A s/ √ n If the test statistic lies beyond the critical value(s) (in the rejection region) we reject H0. We say THERE IS SUFFICIENT EVIDENCE TO REJECT H0. If the test statistic does not lie beyond the critical value, we accept H0. We say THERE IS NOT SUFFICIENT EVIDENCE TO REJECT H0
- 16. Normal distribution X ∼ N(µ, σ2 ) and the theory behind the Z-test and the T-test If samples of size n are taken from a population with mean A and standard deviation σ, then the sample means are distributed normally, with mean A and standard deviation σ/ √ n. −4 −2 2 4 0.1 0.2 0.3 0.4 0.5 x y When we calculate the test statistic, we are calculating the Z-score of the sample mean. The critical value is the Z-score of a sample mean which we have a 5% (or 1%) probability of obtaining. For further information, try a statistics book from the library, or the khanacademy videos on youtube.
- 17. T-test - Example 1 A light bulb company claim their light bulbs last an average of 1000 hours. We want to test whether this is true to a 5% level of signiﬁcance. H0: µ = 1000. H1: µ = 1000. We test a sample of 10 light bulbs. Their lifetimes in hours are listed below. 1020, 860, 987, 1109, 1015, 952, 964, 1007, 1082, 1017 Degrees of freedom:(d.o.f. = n − 1) 10-1=9 Critical values: We are doing a two-tailed test as our alternative hypothesis says µ = 1000. Look up 5% with 9 degrees of freedom for the critical value. Our critical values are -2.26 and 2.26.
- 18. T-test - Example 1 Sample mean: ¯x = 1001.3. Sample standard deviation: s2 = x2 − n¯x2 n − 1 = 4768.9 s = √ 4768.9 = 69.057 Test statistic: ¯x − A s/ √ n = 1001.3 − 1000 69.057/ √ 10 = 0.06 Decision: −2.26 < 0.06 < 2.26 .The test statistic is not in the rejection region so we accept the null hypothesis. Conclusion: The sample of 10 light bulbs does not provide sufﬁcient evidence at a 5% signiﬁcance level to reject the light bulb company’s claim; the average bulb lifetime is 1000 hours
- 19. T-test - Example 2 An average person is said to be able run to 100m in 14.2 seconds. We think that this is a bit on the slow side. We decide to test at a 5% level of signiﬁcance. H0 : µ = 14.2 H1 : µ < 14.2. We ask 7 people to run 100m. Their times are as follows: 12.6, 13.2, 11.7, 14.6, 11.3, 12.0, 13.5 The degree of freedom of this set is 7-1=6 We are doing a one-tailed test as our alternative hypothesis says µ < 14.2. Look up 5% with 6 degrees of freedom for the critical value. The critical value is −1.94.
- 20. T-test - Example 2 Sample mean ¯x = 12.7. Sample standard deviation s2 = x2 − n¯x2 n − 1 = 1.327 s = √ 1.327 = 1.152 Test statistic T = ¯x − A s/ √ n = 12.7 − 14.2 1.152/ √ 7 = −3.45 Decision: −3.45 < −1.94 so we reject the null hypothesis. Conclusion: The data collected provides sufﬁcient evidence at a 5% signiﬁcance level to reject the claim that the average person runs 100m in 14.2s; people run faster than this.
- 21. T-test - Example 3 An average person has an IQ of 100. We think that we are cleverer than this so we test at a 1% level of signiﬁcance. H0 : µ = 100. H1 : µ > 100. We got 8 people to take an IQ test. Their marks were as follows: 117, 106, 93, 142, 110, 114, 120, 126 The degree of freedom of this set is 8-1=7 We are doing a one-tailed test as our alternative hypothesis says µ > 100. Look up 1% with 7 degrees of freedom for the critical value. The critical value is 3.00.
- 22. T-test - Example 3 Sample mean ¯x = 116. Sample standard deviation s2 = x2 − n¯x2 n − 1 = 208.857 s = 14.452 Test statistic ¯x − A s/ √ n = 116 − 100 14.452/ √ 8 = 3.13 Decision: 3.00 < 3.13 so we reject the null hypothesis. Conclusion: The sample of 8 people provides sufﬁcient evidence at a 1% signiﬁcance level to reject the claim that the average IQ is100; people are more intelligent than this.
- 23. T-test - Example 4 The chocolate company claims that a bag of malteasers has an average of 20 malteasers inside. In the name of science we buy 6 bags to see if this is right to a 1% level of signiﬁcance. The bags have the following number of malteasers: 19, 16, 18, 19, 22, 14 H0 : µ = 20. H1 : µ = 20. Degree of freedom is 6-1=5. We are doing a two-tailed test as our alternative hypothesis says µ = 20. Look up 1% with 5 degrees of freedom for the critical values. The critical values are −4.03 and 4.03.
- 24. T-test - Example 4 Sample mean ¯x = 18. Sample standard deviation s2 = x2 − n¯x2 n − 1 = 7.6 s = 2.757 Test statistic T = ¯x − A s/ √ n = 18 − 20 2.757/ √ 6 = −1.78 Decision: −4.03 < −1.78 < 4.03 so we accept the null hypothesis. Conclusion: The sample of 6 bags of maltesers does not provide sufﬁcient evidence at a 1% signiﬁcance level to reject the chocolate company’s claim; there is an average of 18 maltesers per bag.

No public clipboards found for this slide

Be the first to comment