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Tutorial 9

Tutorial 9. Statistics. Normal Distribution (Ex 1).

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Tutorial 9

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  1. Tutorial 9 Statistics

  2. Normal Distribution (Ex 1) • An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed, with mean equals to 800 hours and a standard deviation of 40 hours. Find the probability that random sample of 16 bulbs will have an average life less than 775 hours.

  3. Ex 1 (con’t) • Solution: • By central limit theorem, is normally distributed with μ=800 and • Corresponding to = 775, we find that Z = (775-800)/10 = -2.5Then,

  4. Confidence Interval • Since table for is not readily available, we transform the distribution to N(0,1) by 1-a a/2 a/2 za/2 z1-a/2 0

  5. Confidence Interval

  6. Confidence Interval (Ex 2) • An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed, with a standard deviation of 40 hours. If a random sample of 30 bulbs has an average life of 780 hours, find a 96% confidence interval for the population mean of all bulbs produced by this firm.

  7. Ex 2 (con’t) • Solution: • 96% => α= 0.04 • Then 96% confidence interval is hence given as • From the table, we have Z0.98 =2.055, thus

  8. Significance Testing • Step 1. State the null hypothesis, H0 and the alternative hypothesis. • Step 2. Consider the appropriate distribution given by the null hypothesis. • Step 3. Decide on the level of the test. This fixes the critical values of the test statistic. • Step 4. Decide on the rejection criteria • Step 5. Calculate the value of the statistic • Step 6. Make conclusion.

  9. Significance Testing (Ex 3) • The r.v. X is such that X ~ N(μ,100). A value is taken at random from the population and found to be 172. Test, at the 5% level, whether the population mean μ could be 150.

  10. Ex 3 (con’t) Solution: • Step 1: We assume that μis 150. So, H0 : μ = 150 H1 : μ not equal 150 • Step 2:Now if H0 is true, X ~ N(150,100) • Step 3 & 4:For 5% level, where Z = (X -μ)/σ, we will reject H0 if |Z| > 1.96

  11. Ex 3 (con’t) • Step 5.Now z = (x-μ)/σ = (172-150)/10 = 2.2 • Step 6.Since |z| > 1.96, we reject H0 and conclude that there is significant evidence, at the 5% level, to suggest that the population mean is not 150.

  12. Type I and II Error

  13. Type I and II Error (Ex 4) • To test whether a coin is fair, the following decision rule is adopted. Toss the coin 120 times; if the number of heads is between 50 and 70 inclusive, accept the hypothesis that the coin is fair, otherwise reject it. • a) Find the probability of rejecting the hypothesis when it is correct. (Type I error) • b) With the original decision rule, find P(Type II error) if the coin is biased and the probability that a head is in fact 0.6

  14. Ex 4 (con’t) Solution: • Let X be the r.v. ‘the number of heads obtained’. • Then X~Bin(n,p) with n=120 • Now, since n is large, X ~ N(np, npq)* approximately, where q = 1-p. • Under H0, np = 120x(1/2) = 60 npq= 120x(1/2)x(1/2) = 30So X~N(60,30) • *(Remark: we can estimate the r.v. by normal distribution when n>30)

  15. Ex 4 (con’t) a) Type I error

  16. Ex 4 (con’t) b) Type II error • We accept H0 if the number of heads lies between 50 and 70 inclusive. • Now P(Type II error) = P(accepting H0|H1 is true) = P(49.5 < X < 70.5 |p =0.6) • Now, if p=0.6, np = (120)(0.6) = 72 npq = (120)(0.6)(0.4) = 28.8

  17. Ex 4 (con’t) • Under H1, X~N(72, 28.8) • Therefore, P(Type II error) = 0.390 (3 sig)

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