Does Smoking Have An Effect On Your Mental Ability?
...elation. Here is my sample of 80 for both smokers and non-smokers: Number IQ Results for Non-Smoker IQ Results for Smoker 1 150 138 2 145 143 3 185 178 4 160 143 5 160 163 6 145 173 7 160 153 8 160 143 9 155 153 10 165 133 11 155 163 12 185 148 13 180 158 14 175 123 15 170 143 16 170 143 17 160 153 18 150 128 19 170 148 20 180 143 21 155 185 22 130 138 23 170 153 24 155 143 25 165 128 26 170 133 27 170 128 28 160 123 29 155 153 30 150 123 31 205 163 32 170 138 33 160 143 34 150 133 35 155 138 36 185 168 37 165 123 38 180 148 39 160 158 40 180 148 41 170 148 42 210 143 43 165 153 44 190 118 45 180 128 46 145 123 47 165 143 48 165 143 49 165 148 50 175 128 51 190 153 52 160 163 53 180 138 54 170 123 55 175 143 56 180 148 57 155 158 58 160 148 59 180 143 60 155 158 61 175 163 62 145 123 63 150 123 64 200 148 65 160 128 66 180 153 67 135 173 68 130 158 69 160 133 70 155 123 71 175 143 72 170 128 73 160 138 74 175 158 75 160 143 76 160 148 77 160 128 78 125 123 79 130 138 80 185 143 You can already see that the scores of smokers are considerably lower than that of non-smokers. I will now use the Central limits theorem to produce a confidence interval of 95%. The Central limits theorem is useful as it allows you to make predictions about a distribution of a sample mean even if you do not know about the parent population. It is vital to have a good cross-section of the data for the central limits theorem to work. N = Number of sample = 80 σ = Standard Deviation σ 2 = Variance µ = Mean σ/√ n = Standard error (s.e.) x = Sample Mean s2 = Sample Variance Standard error is the term used to define the standard deviation of the distribution of the sample means. Non- Smokers First to continue with my work I need to work out the sample variance using the formula: ∑(x2) _ ∑x2 n n Meaning; the average of the squares minus the square of the average. So, S2 = 27902.92 – 166.41666672 = 208.4097 The sample mean is an unbiased estimator of the population mean. I know my sample mean, however I do not know my original variance for my parent population, to find this out I will use an unbiased estimator in the form: n n – 1 x s2 = σ2 and s2 = 208.4097 Therefore: 100 99 x 208.4097 = 210.5148485 = σ2 σ = 14.509 so 14.5 An unbiased estimator is a quantity that does not show any bias and I have used one so that my work is not affected by any bias, therefore making the test fair. This is my parent population for non-smokers which I will take my sample from and standardise to form a normal distribution. Now I will do a confidence interval of 95% using my sample to further express the degree of my estimate. 14.5 14.5 µ - 1.96 x √ 60 < x < µ + 1.96 x √ 60 x = 166.4167 Therefore: 14.5 14.5 166.4167 - 1.96 x √ 60 < µ < 166.4167+ 1.96 x √ 60 162.75 < µ < 170.09 so about 163 < µ < 170 I am 95% confident that all of my sample means for non-smokers are between 163 and 170. Here’s how I got the figure 1.96: If you look at this distribution, it shows a 95% interval, therefore there is 5% left over. 2.5% in each side. So Φ(z) = 1 - 0.025 = 0.975 this figure is the used to look at the normal distribution to see what z equals. The normal distribution shows that it is 1.96. I will now repeat the last few steps for smokers. Smokers S2 = 20937.58 – 144.08333332 = 177.67 100 σ2 = 99 x 177.67 = 179.464 σ = 13.4 Now the 95% confidence interval: 13.4 13.4 µ - 1.96 x √ 60 < x < µ + 1.96 x √ 60 x = 144.083 Therefore: 13.4 13.4 144.083 - 1.96 x √ 60 < µ < 144.083 + 1.96 x √ 60 140.69 < µ < 147.4 so about 141 < µ < 147 I am 95% confident that all of my sample means for smokers are between 141 and 147. Conclusion My results show me that the IQ of smokers is lower that that of non-smokers. The data shows that the peak of smokers is around 175, and non-smokers around 210. This shows a great difference. These results suggest that non-smokers have a hig...