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STAT131 W8L2 Sampling Distributions Z, t

STAT131 W8L2 Sampling Distributions Z, t. by Anne Porter alp@uow.edu.au. Lecture Outline. Review continued from W8L1 Sampling distributions Problems using Z (Normal, Gaussian) Probability re Individual observations Probability re means Problems using T (T distribution)

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STAT131 W8L2 Sampling Distributions Z, t

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  1. STAT131W8L2 Sampling Distributions Z, t by Anne Porter alp@uow.edu.au

  2. Lecture Outline • Review continued from W8L1 • Sampling distributions • Problems using Z (Normal, Gaussian) • Probability re Individual observations • Probability re means • Problems using T (T distribution) • Variance unknown, small sample but from a normal population • Variance unknown, large population

  3. Sampling Distributions • The distribution of all possible values that can be assumed by some statistic (eg mean, variance), computed from samples of the same size randomly drawn from the same population, is called a sampling distribution of that statistic. • We can construct these distributions • by exhaustively taking all samples of a given size • by simulation taking a large number of samples

  4. Method of sampling • With replacement or • Without replacement • The choice of method will result in a different distribution. • Sometimes the choice will make little difference • eg if the sample is small in relation to a large population

  5. Central Limit Theorem(Large Sample Normality) • Given a random sample X1, X2, ..Xn from any distribution with mean m and finite variance s2, then irrespective of the distribution of the parent population, the distribution of approaches the shape of a normal distribution when sample size is large, with a mean and standard deviation and irrespective of sample size.

  6. Example 1 about single scores • A study of a large group of 8 year old female students at a state primary school has shown that they have an average IQ of 100 with a standard deviation of 10. (i) What is the probability that a girl selected at random from this group has an IQ of over 105?

  7. Example 1 about single scores • A study of a large group of 8 year old female students at a state primary school has shown that they have an average IQ of 100 with a standard deviation of 10. (i) What is the probability that a girl selected at random from this group has an IQ of over 105? This asks for P(X>105) and to do this we need to standardise x and this is equivalent to P(Z>(x-m)/sIn this case we have P(Z>(105-100)/10) that is P(Z>0.5)

  8. P(Z>0.5) Z=0.5 Use a Diagram Equivalent to X=105 And consult the Z tables

  9. Consult Normal tables Find P(Z>0.5)

  10. Consult Normal tables Find Z=0.5 Then check second decimal place Z=0.50 Then read the probability of Z <z Then P(Z>0.5)=1-0.6915=0.3085

  11. Example 2 about means • A study of a large group of 8 year old female students at a state primary school has shown that they have an average IQ of 100 with a standard deviation of 10. (ii) If we take sample of 5 students what is the probability that the mean of the IQs is greater than 105?

  12. Here we have a sample mean so to standardize , we will use the central limit theorem, Example 2 about means • A study of a large group of 8 year old female students at a state primary school has shown that they have an average IQ of 100 with a standard deviation of 10. (ii) If we take sample of 5 students what is the probability that the mean of the IQs is greater than 105?

  13. P(Z>1.12) Use a Diagram Equivalent to And consult the Z tables

  14. Consult Normal tables Find Z=1.12 Then check second decimal place Z=1.12 Then read the probability of Z <z Then P(Z>1.12)=1-0.8686=.1314

  15. Example 3 • A study of a large group of 8 year old female students at a state primary school has shown that they have an average IQ of 100 with a standard deviation of 10. (iii) What is the probability that the first girl has an IQ of over 105 and the second girl has an IQ of under 100?

  16. Example 3 • A study of a large group of 8 year old female students at a state primary school has shown that they have an average IQ of 100 with a standard deviation of 10. (iii) What is the probability that the first girl has an IQ of over 105 and the second girl has an IQ of under 100? Recognising these as independent events, knowing P(X>105)=.3085 (example1) and realising 100 is the mean ie P(X<100)=0.5 we have using multiplication rule for two independent events P(X>105)xP(X<100)= 0.3085x0.5

  17. Ex 4: Mean & Variance Exponential If samples of size 49 are simulated from the exponential distribution with a rate lambda of 1/3. • What is the variance and standard deviation of the population? Population distribution

  18. Ex 4: Mean & Variance Exponential If samples of size 49 are simulated from the exponential distribution with a rate lambda of 1/3. • What is the variance and standard deviation of the exponential population? For an exponential So =3 or m=s=1/l So s2=Variance=9

  19. Ex 5 Central Limit Theorem -shape If samples of size 49 are simulated from the exponential distribution with a rate lambda of 1/3. • What will be the shape of the distribution of the means of the samples? Distribution of population scores

  20. Normal or bell shaped Sampling distribution of means Ex 5 Central Limit Theorem -shape If samples of size 49 are simulated from the exponential distribution with a rate lambda of 1/3. • What will be the shape of the distribution of the means of the samples? Distribution of population scores

  21. Ex 5 Central Limit Theorem - mean If samples of size 49 are simulated from the exponential distribution with a rate lambda of 1/3. • What will be the mean of the of the sampling distribution of the mean?

  22. Ex 5 Central Limit Theorem - mean If samples of size 49 are simulated from the exponential distribution with a rate lambda of 1/3. • What will be the mean of the of the sampling distribution of the mean? From the central limit theorem the mean of the sampling distribution equals the mean of the population The mean of an exponential is m=1/l therefore and therefore

  23. Ex 5: Central Limit theorem - variance If samples of size 49 are simulated from the exponential distribution with a rate lambda of 1/3. • What is the variance and standard deviation of the sampling distribution of means? Given we had the population standard deviation for the exponential

  24. Ex 5: Central Limit theorem - variance If samples of size 49 are simulated from the exponential distribution with a rate lambda of 1/3. • What is the variance and standard deviation of the sampling distribution of means? From the central limit theorem the standard deviation of the sampling distribution is given by Hence the variance is 0.4292

  25. Relative frequency 0.14 0 2 4 6 8 10 12 Population Values Reminder: • Even with a population shaped like this uniform discrete population when we plot the distribution of the means of all samples of size two we get….next slide

  26. rel freq 0.14 0.12 0.10 0.08 0.06 0.04 0.02 0 0 1 2 3 4 5 6 7 8 9 10 11 12 Means of samples Size 2 Symmetric, discrete (not bell shaped but is moving that direction from the original population). As the sample size increases The distribution of the sample means is... we would expect this this to become closer to a normal curve

  27. Sampling Distribution of the Means unknown – T distribution • If is the mean of a random sample of size n taken from a normal population having a mean m and, unknown variance s2, estimated by then t= • T is a random variable having the t distribution with the parameter v=n-1 as degrees of freedom..

  28. Example 6 : - t distribution A manufacturer of fuses claims that with a 20% overload, the fuses will blow in 12.40 minutes on average. To test this claim, a sample of 20 of the fuses was subjected to a 20% overload, and the times it took to them to blow had a mean of 10.63 minutes and a standard deviation of 2.48 minutes. If it can be assumed that the data constitute a random sample from a normal population, do they tend to support or refute the manufacturer's claim?

  29. When the variance of the population is unknown • And either the sample is large OR it is reasonable to assume the sample comes from a normal population • We use the statistic • With n=n-1 degrees of freedom • Hence we calculate (convert to t)

  30. Example 6 : - t distribution Fuses will blow in 12.40 minutes on average. Test this claim. For a sample of 20, the times it took to them to blow had a mean of 10.63 minutes and a standard deviation of 2.48 minutes. The data constitute a random sample from a normal population, do they tend to support or refute the manufacturer's claim?

  31. Example 6 : - t distribution Fuses will blow in 12.40 minutes on average. Test this claim. For a sample of 20, the times it took to them to blow had a mean of 10.63 minutes and a standard deviation of 2.48 minutes. The data constitute a random sample from a normal population, do they tend to support or refute the manufacturer's claim? • With n=n-1 =20-1=19 degrees of freedom • To find (approximate) the probability of getting a value this extreme consult the t -tables

  32. So p of 3.218 is less than .005 p tp 0 3.218 When the variance of the population is unknown • With n=n-1 =20-1=19 degrees of freedom and • What is the probability of getting a as extreme as t=- 3.218 and given a symmetric distribution as extreme as 3.218

  33. Next lecture • Confidence Intervals using Z and T distributions

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