Chapter 7- The Single Sample t-Test
Chapter 7- The Single Sample t-Test Psy 202
Popular in Elementary Statistics
Popular in Psychology (PSYC)
This 3 page Class Notes was uploaded by T'Keyah Jones on Thursday September 29, 2016. The Class Notes belongs to Psy 202 at University of Mississippi taught by Dr. Melinda Redding in Fall 2016. Since its upload, it has received 6 views. For similar materials see Elementary Statistics in Psychology (PSYC) at University of Mississippi.
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Date Created: 09/29/16
September 29, 2016 a) Single Sample t-Test Used to compare a sample mean to a population mean when the standard deviation of the population is unknown The steps are the same used for the z-test b) Going from “z” to “t” To compute a z-score, the population standard deviation (or variance) must be known. Estimated Standard Error The sample deviation is substituted for the population standard deviation in the calculation Using this in the denominator of the test statistics led to the t-distribution S M Estimated standard error of means Formula: _S_ √N M Standard error of means c) t-Statistics Used to determine the number of standard deviations in a t-distribution that a sample deviates from the mean value or differences stated in the null The t-distribution is identical to the z-distribution Small sample sizes equal larger critical values t= M - μ S √N d) Degrees of Freedom Symbol: df Formula: N-1 As the df of the sample variance increases, the shape of the t-distribution changes As the sample size increases, the sample variance more closely resembles the population variance To Know Critical Values 1. Know the degree of freedom 2. Determine the type of tail (One-tailed or Two-tailed) 3. Know the alpha level e) Examples M= 62,000 N= 16 μ= 66,000 S= 12,000 Step 1: Determine Type of Test Single sample t-test because we want to compare the mean from a sample with an unknown standard deviation Step 2: Assumptions 1. Random sample 2. Independent Observation 3. Normal distribution (dependent variable) Step 3: Hypotheses H 0 μFISH 66,000 ; pollution does not have an effect H 1 μFISH 66,000 ; pollution does have an effect Step 4: Decision Rule = .05 df= 15 (16-1) Critical value: 1.753 (One-tail test on chart) Step 5: Calculations t= M - μ t= 62,000 - 66,000 = -1.333 S_ 12,000 √N √16 Step 6: Interpret We fail to reject the null because -1.333 is not in the tail of the distribution; there is not enough evidence to suggest that the polluted lakes have significantly fewer fish Type II error Example 2 N= 14 M= 42 S= 6.8 μ= 36 Step 1 t-test because we want to compare the mean from a sample to the mean of the general population with an unknown standard deviation Step 2 Refer to 1 example Step 3 H 0 μ KILLERS6; there is no significant difference in introversion scores H 1 μKILLERS36; there is a significant difference in introversion scores Step 4 = .05 df= 13 (14-1) Critical value: 2.160 (Two-tailed) Step 5 t= M - μ t= 42 -36 = 3.302 S_ 6.8 √N √14 Step 6 We will reject the null because 3.302 is in the tail of the distribution; serial killers have significantly different introversion scores than the general population
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