Significance Tests our response:

1. UK data for male median earnings indicates that this figure is £ 24 000. What can you say about the validity this figure if a simple random sample of 200 families in Britain showed an average earnings level of £ 23 500, with a standard deviation of £ 4000? Use a .05 level of significance. Would your conclusions be any different at a .01 level of significance?
   
   The claimed mean income is £ 24 000
   The sample mean income is £ 23 500 on a sample of 200, with a standard deviation of £ 4000.
   
   To calculate the standard error:
   
   = standard deviation / square root of the number in the sample
   = 4000 / 200
   Standard Error = 283
   The difference between the sample and the claimed mean income = - 500
   This difference, divided by the standard error (- 500 / 283) = - 1.77
   
   This z - score of - 1.77 is within both +/- 1.96 at 0.05 probability, and +/-2.58 at 0.01 probability.
   
   This demonstrates the validity of the claimed median earnings figure at both significance levels. The null hypothesis would be accepted.
   
   

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2. A manufacturer of plasticised line used in home-assembly mobiles advertises that their product has an average tensile strength of 30 kilograms. You took a sample of 100 sections of the line and tested them. The average tensile strength of this sample was 28 kilos, with a standard deviation of 12 kilos. Does this enable you to dismiss the manufacturer's claims? Answer this question for a significance level of .05 and .01. Think about whether the more appropriate test will be one-tailed or two-tailed.
   
   The claimed tensile strength of the line = 30 kgs
   The sample mean strength = 28 kgs on a sample of 100, with a standard deviation of 12 kgs.
   
   Should the test be one or two tailed?
   A one-tailed test should be used because the test is to determine if the actual strength of the line is less than that claimed by the manufacturer.
   
   At .05 significance level, the z-score is +/- 1.64
   At .01 significance level, the z-score is +/- 2.33
   
   Standard error = standard deviation / square root of the number in the sample
   = 12 / 10 = 1.2
   
   The difference between the sample mean and the claimed mean = - 2 kgs
   This produces a z-score of - 2 / 1.2 = - 1.67
   
   A z-score of - 1.67 is greater than - 1.64, so there is only a .025 probability that the sample comes from the same distribution as claimed. Therefore at a significance level of .05, we can say that there is a 95 % certainty that the sample is from a different population.
   
   At a .05 significance level we would reject the null hypothesis and accept the alternative hypothesis.
   
   But at a significance level of .01, - 1.67 is less than - 2.33, so there is only a .01 probability that the sample comes from a different distribution than claimed. So, there is a 99 % certainty that the sample is from the same population as is claimed by the manufacturer.
   
   Here, at a .01 significance level we would accept the null hypothesis.
   
   

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3. Self-test on comparing two sample means: Using a z-test.
   
   The difference between the two sample means is 2.4, in favour of Lecturer 2. Is this significant or just a chance result arising from the two samples. You have to set up the Null and Alternative Hypotheses as follows:
   
   Null H₀ : There is no real difference in the results. Both groups of results come from the same population. The calculated difference is insignificant and the students' complaints are unjustified.
   
   Alternative H₁ : A real difference exists. The two sample means come from two different populations. The complaints are justified.
   
   Standard Error = standard deviation of the sample / square root of the number in the sample. There are 30 observations in each sample. Square root of 30 = 5.48.
   
   Lecturer 1: Standard Error = 7.2 / 5.48 = 1.31
   Lecturer 2: Standard Error = 8.4 / 5.48 = 1.53
   
   Standard Error (differences) = Square Root {(Standard Error Group 1)² + (Standard Error Group 2)²}
   
   Standard Error (differences) = Sq Rt (1.31² + 1.53²)
   Standard Error (differences) = Sq Rt (1.72 + 2.34)
   Standard Error (differences) = 4.06 = 2.01
   
   Now, the difference between the two sample means was 2.4.
   The z-score for this difference = 2.4 / 2.01 = 1.19
   This value falls between 1 and 2 standard deviations from the mean.
   
   A z-score of 1 includes approximately 68 % of all possible scores. There is, in other words a .68 probability that the result will fall within a z-score of +1 and -1. Alternatively, there is a .32 probability that any result will fall beyond a z-score of +/- 1 on either side of the mean.
   
   The figure obtained for the difference between the two means here is a z-score of 1.19. We can say with 68 % certainty that the two samples come from different populations. Therefore the difference is significant and the two lecturers' results indicate inconsistent standards.
   
   But a confidence level of .32 is not very stringent. If we set the level at .05, we cannot reject the Null Hypothesis because a z-score of 1.19 falls well within the limits indicating a .05 level of certainty (1.96).
   
   If we want to be 95 % sure of our conclusions, we must agree that there is no difference between the standards of the two lecturers. Therefore, the students' complaints are unjustified.