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## Outline

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**Outline**• Review of last week • Sampling distributions • The sampling distribution of the mean • The Central Limit Theorem • Confidence intervals • Normal distribution example • Sampling distribution example • Confidence interval example**Review of last week**• Last week, we learned how to use the Standard Normal Distribution to work out the probability of finding individual scores in some interval – e.g., what is the probability that the next Canadian woman we meet is taller than 175 cm? • Today, we’re going to do the same sort of thing with sample meansrather than individual scores.**The sampling distribution of a sample statistic (such as X)**is the probability distribution of that statistic. Population (µ) Sample X**The sampling distribution of the mean consists of all**possible sample means – for all possible samples of size n – that you could take from the population Population (µ) Sample 4 X4 Sample 3 X3 Sample 2 X2 Sample 1 X1**When we draw a sample from a population, we are at the same**time drawing a sample mean from the distribution of sample means for samples of size n X Distribution of sample means for samples of size n µX**The sampling distribution of a sample statistic is the**probability distribution of that statistic. We can have sampling distributions of any sample statistic Mean Median M Variance s2 Std devn s Sampling distributions**The sampling distribution of the mean**• The sampling distribution of the sample mean X. E(X) = μ = μ • Variability of this distribution is given by the standard error of the mean: σ = σ ≅ s**Consider a random sample of n observations from a population**with mean µ and standard deviation . When n is sufficiently large, the sampling distribution of X will be approximately normal with mean µ = µ and = / . Note: this is true regardless of the shape of the underlying distribution of raw scores The Central Limit Theorem**The larger the sample size, the better the approximation to**the normal distribution. For most populations, n ≥ 30 will be “sufficiently large.” The Central Limit Theorem**The Central Limit Theorem**• When we draw a sample and measure its mean, by the CLT, we may assume the sampling distribution of the sample mean is normal. • That means we can use the standard normal distribution (SND) to work out the probability of finding a sample mean in a given range relative to the population mean.****μ The sampling distribution of the sample mean**The sampling distribution of the mean**• We use the sampling distribution of the mean the way we used the SND last week. We obtain probabilities of finding sample means in a given range relative to the population mean, for samples of size n. • Don’t forget to use the standard error, σX, rather than the standard deviation, σ!**Confidence Intervals**• There are two ways to estimate population parameters such as the mean: • Point estimates, such as X • Interval estimates, which tell us a range of values that will contain the parameter with known probability.**.45**.45 Z = -1.645 µX Z = 1.645 90% of the time, X will fall within the range Z = -1.645 to Z = +1.645**Confidence Intervals**• If 90% of the time X falls in the range Z = -1.645 to Z = +1.645 around the mean µ, then… • 90% of the time, µ must fall within a range of the same width centered on X.**Confidence Intervals**• For given , the 100 (1-)% Confidence Interval for µX is: • C.I. = X ± Z/2 X • C.I. = X ± Z/2 /√n**Confidence Intervals**• When is not known and n is large (≥ 30), use s: • C.I. = X ± Z/2 sX • C.I. = X ± Z/2 s/√n**Normal Distribution Example**• The amount of time that students wait to be served when buying coffee from the “Campus Perks” coffee outlet is normally distributed with a mean of 62.0 seconds and a 98.5 percentile of 79.36 seconds. In a random sample of 30 students buying coffee at Campus Perks, approximately how many will wait between 40 and 58 seconds to be served? NOTE: This is not a question about a sample mean!**Normal Distribution Example**.50 .4850 40 58 62 P98.5 Z for .4850 = 2.17**Normal Distribution Example**• = 79.36 – 62 = 8 • 2.17 • Z1 = 40 – 62 = -2.75 (p = .4970 from table) • 8 • Z2 = 58 – 62 = -0.50 (p = .1915 from table) • 8**Normal Distribution Example**• P(40 ≤ X ≤ 58) = .4970 - .1915 = .3055 • The probability of any one student waiting between 40 and 58 seconds is .3055. • Therefore, in a random sample of 30, we expect approximately .3055 (30) = 9.165 ≈ 9 students to wait between 40 and 58 seconds.**Sampling Distribution Example**• People’s reaction times (RTs) to a simple visual stimulus are normally distributed with a mean of 500 milliseconds and a standard deviation of 150 milliseconds. You believe that people who go on a low-carb diet, however, will have slower (longer) RTs than this, on average, though their standard deviation will remain at 150. To test your belief, you take a random sample of 40 people who self-report having being on a low-carb diet for at least 6 months and measure their RTs. You decide that your belief will be supported if the mean RT of the low-carb group is 565 milliseconds or slower. What is the probability that you will conclude that your belief has been supported even if a low-carb diet actually has no effect on RTs whatsoever?**We want this probability**500 565 You decide that your belief will be supported if the mean RT of the low-carb group is 565 milliseconds or slower. What is the probability that you will conclude that your belief has been supported even if a low-carb diet actually has no effect on RTs whatsoever?**Example 2**• What is P(X ≥ 565 │µ = 500)? • Z = 565 – 500 • 150/√40**Example 2**• What is P(X ≥ 565 │µ = 500)? • Z = 565 – 500 = 65 = 2.74 • 150/√40 23.72 • P for Z = 2.74 (from table) is .4969. • Therefore, desired probability is .5 - .4969 = .0031.**Example 3**• Two variables important to a professional football player are speed and strength. Each year, camps are held to determine potential players’ speed and strength, both of which are continuous, normally-distributed, and independent of each other. The middle 95% of strength scores is bounded by 600 and 900 (on a composite strength index). The average time to run 40 yards is 4.6 seconds, and 40 yard time exceeds 6 seconds only 5% of the time. • a. In order to be considered by a team, a potential player must not exceed the 75th percentile for time to run 40 yards. What is the slowest a player can run 40 yards and still be considered?**.25**4.6 6 .45 seconds X Probability distribution for time to run 40 yards (seconds)**Example 3**• Z(.45) = 1.645 = 6 – 4.6 • σ • σ = 6 – 4.6 = .851 • 1.645**Example 3**• Now we can find X (the 75th percentile): • Z(.25) = 0.675 = X – 4.6 • .851 • X = 0.675 * (.851) + 4.6 = 5.15 (seconds)**4.6**6 seconds 5.15 The 75th percentile for 40 yard times is 5.15 seconds.**Example 3**• Two variables important to a professional football player are speed and strength. Each year, camps are held to determine potential players’ speed and strength, both of which are continuous, normally-distributed, and independent of each other. The middle 95% of strength scores is bounded by 600 and 900 (on a composite strength index). The average time to run 40 yards is 4.6 seconds, and 40 yard time exceeds 6 seconds only 5% of the time. • b. You take a random sample of 200 potential players. What is the probability that the average strength score of the sample is less than or equal to 740?**.45**.45 600 900 µ 750 Probability distribution for strength scores**Example 3**• Z = 1.645 = 900 – 750 • σ • σ = 900 – 750 = 91.19 • 1.645 • Z = 740 – 750 = -1.55 • 91.19/√200**Example 3**• P (Z < 1.55) = .4394 (From table) • Tail probability will be .5 – .4394 = .0606**This is the sampling distribution of mean strength scores**for samples with n = 200 .0606 740 What is the probability that the mean for a sample of 200 players is less than this value? 750**Confidence Interval Example**• A researcher samples 36 undergraduates from a local university and finds it took them 36.4 days, on average, to find a job, with a standard deviation of 8 days. Use these data to form a 96% confidence interval for the true mean time it takes for graduates to find a job. NOTE: We are not given the population standard deviation**Confidence Interval Example**• Recall: • C.I. = X ± Z/2 sX = X ± Z/2 s/√n • X = 36.4 • S = 8 • n = 36 • S/√n = 8/6 = 1.33**Confidence Interval Example**• (1-)% = 96%, so /2 = .02 – this is the tail probability. • We get /2 = .02 when we look up Z.48 = 2.05 • C.I. = 36.4 ± 2.05 (1.33) • (33.67 ≤ µ ≤ 39.13)