Chapter 12 Sampling Distributions The Sampling Distribution of the Mean The Normal Deviate Test
Quiz Beep Beep believes that he can influence coins so that they land heads more often than tails. His null hypothesis is that the coins land heads 50% of the time or less. If he wants to be 95% sure he is correct if he rejects the null hypothesis H0, what is his alpha level?
Sampling Distribution of the Mean The average penny is about 16 years old with a standard deviation of 15. μ = 16 σ = 15 Suppose we take a sample of N coins (say 30 coins) and calculate the average of the sample, Sample mean = If we take lots of samples, what do you think the mean of all these would be? μ = ? (mean of the sampling distribution of the mean)
Sampling Distribution of the Mean. μ = μ = 16 years in our case. What would be the standard deviation of all these means? σ = ? (standard deviation of the sampling distribution of the mean) σ = σ/ √ N = 15/ √ 30 = 2.7 years If you had a penny that was 25 years old, could you tell if it came from somewhere different than the population in question?
Intro to the z-test If you had a penny that was 25 years old, could you tell if it came from a different population than the population in question (i.e., someone’s coin collection)? μ = 16 σ = 15 Calculate z-score: z = (x – μ) / σ z = (25 -16)/ 15 z = -0.6 => nothing unusual => Retain the null hypothesis. => This penny might have been from the same population.
Intro to the z-test If you had a sample of 30 pennies that with an average age of 25 years (= obt), could you tell if it came from a different population from the population in question? μ = 16, μ = 16 σ = 15, σ = σ / √ N = 15 / √ 30 = 2.7 Calculate z-statistic obtained from the sample zobt: zobt = ( obt – μ ) / (σ/ √N) zobt = (25 -16)/ (15/ √30) zobt = -3.3 => Extremely low z-statistic => Reject the null hypothesis. => These pennies came from somewhere else.
The Normal Deviate Test (z-test) A test to see if a sample is different from a given population. We calculate a z statistic (zobt) for the sample to calculate the probability of getting a sample with that mean from the population. Mean = µ Standard deviation = σ = σ/√N If the probability is low (a high |zobt|), we say that the sample is not likely to have come from the given population.
College Group Example Your college group at church has an average attendance of 100 with a standard deviation of 10. You have a big outreach activity at the end of April. In May, the average of the 4 Sundays’ attendance is 105. (specifically, 100, 105, 95, 120) Does this indicate that the group has grown? Can you be 95% sure that the group has grown in May?
College Group Example Great Commission Model of Outreach “Go and make disciples out of all nations” Matt. 28:19 Outreach Evangelism Discipleship
College group example Step 1: Define the alternate and null hypotheses H1: The college group has changed size (non-directional, 2 tailed). SizeAfter ≠ SizeBefore H0: The college group has remained the same size. SizeAfter = SizeBefore This will be retained if it is reasonable to consider that the sample of 4 scores with obt = 105 is a random sample from a population with μ = 100 and σ =10.
College group example Step 2: Calculate the statistic we need. If we have 1) the mean and the standard deviation of the population and 2) a single sample mean, we can use a z-test We need zobt. z obt = ( obt– μ)/(σ/√N) μ = 100, σ =10 (Population) obt = 105, N = 4 (Sample) z obt = (105-100)/(10/√4) = 1.00
College group example Step 3: Find the probability of obtaining the statistic if H0 is true; reject H0 if the probability is too low. zobt = 1.00: We want to find the probability of getting results this extreme or more extreme by chance (2 tailed). Using Excel, the area to the left of zobt = 1- normsdist(1) Area to left = p = .1587 Since it’s two-tailed, we need to add both tails, so p = .1587 +.1587 = .3174 This is greater than α = .05, so we retain H0. We conclude that the college group might not have actually grown. There’s a 32% chance we would obtain a result this extreme by chance if the null hypothesis were true.
Z-test on Excel. For a one tailed z test, we need to know the: The sample data = Array The population mean = X The population standard deviation = Sigma We use the command ZTEST with the arguments (Array, X, Sigma) to calculate the p value. In this example, p = ZTEST(b3..b6,f3,f4)
z-test on Excel If the ZTEST function gives a value greater than p = .50, use 1 – ZTEST for the correct 1 tail p value. This would occur when you’ve got a negative zobt. If z is negative, we want area in the tail to the left of zobt.
Reporting the Results APA Style The long term average attendance in a college group (M = 100, SD = 10) was compared to the average of a sample (N =4, M = 105) after an outreach activity. Although the sample attendance was larger than the long term attendance (difference = 5), this difference was not significant (z = 1.00, p = .32, 2 tails).
Zcrit (2 tails) A shortcut so you don’t always have to calculate the probability of a z-statistic. With α2-tailed = .05, we look up a value of z such that 5% of the scores are more extreme than this value (2.5% will be on each tail) zcrit = 1.96 for α2-tailed = .05 If |zobt| (the absolute value of zobt) > zcrit we reject H0. |1.00| < 1.96, so we retain H0. The group might not have grown.
Critical Regions (1 tail) zcrit = 1.645 for α1-tailed = .05
College group example Suppose the group averages 105 for 4 months (16 weeks). Can we say it has changed size?
When Can You Use the Normal Deviate Test (z Test)? When you know μ and σ of the population. AND You have a sample (a set of N scores) you want to test to see if it comes from the above population.
Beanie Baby Exercise: Are Dogs Cuter than the Average Beanie? Formulate alternative hypothesis (1 or 2 tail?). Formulate null hypothesis. Randomly select a sample of n dogs and calculate obt. Note that μ (= 5.50) and σ (= 2.5) for the population. Calculate zobt. Calculate the p value of zobt. Reject or retain the null hypothesis.
Are Dogs Cuter than the Average Beanie Baby?