Distribution of a Difference in Means Section 6.4-D Distribution of a Difference in Means

Math SAT by Gender The average math SAT score for males is 534 with a standard deviation of 118. The average math SAT score for females is 500 with a standard deviation of 112. If we were to take random samples of 50 males and 50 females and record the difference in means 𝑥 𝑀 − 𝑥 𝐹 , what shape would the sampling distribution have? Symmetric and bell-shaped Skewed to the right Skewed to the left None of the above Impossible to determine

Math SAT by Gender The average math SAT score for males is 534 with a standard deviation of 118. The average math SAT score for females is 500 with a standard deviation of 112. If we were to take random samples of 50 males and 50 females and record the difference in means 𝑥 𝑀 − 𝑥 𝐹 , where would the sampling distribution be located? 534 500 34 23 18 𝑥 𝑀 − 𝑥 𝐹 =534−500=34

Math SAT by Gender The average math SAT score for males is 534 with a standard deviation of 118. The average math SAT score for females is 500 with a standard deviation of 112. If we were to take random samples of 50 males and 50 females and record the difference in means 𝑥 𝑀 − 𝑥 𝐹 , what will the standard error be? 534 500 34 23 18 𝑥 𝑀 − 𝑥 𝐹 ~ 𝑁 𝜇 𝑀 − 𝜇 𝐹 , 𝜎 𝑀 2 𝑛 𝑀 + 𝜎 𝐹 2 𝑛 𝐹 ~ 𝑁 534−500, 118 2 50 + 112 2 50 ~ 𝑁 34, 23

Confidence Interval for a Difference in Means Section 6.4-CI Confidence Interval for a Difference in Means

Video Games and GPA 210 first-year college students were randomly assigned roommates For the 78 students assigned to roommates who brought a videogame to college: average GPA after the first semester was 2.84, with a sd of 0.669. For the 132 students assigned to roommates who did not bring a videogame to college, average GPA after the first semester was 3.105, with a sd of 0.625. How much does getting assigned a roommate who brings a videogame to college affect your first semester GPA? Give a 90% confidence interval.

Video Games and GPA 78 students whose roommates brought a videogame to college: average GPA was 2.84, with a sd of 0.669. 132 students whose roommates did not bring a videogame to college: average GPA was 3.105, with a sd of 0.625. For a 90% confidence interval, what is t*? a). 1.645 b). 1.960 c). 2.576 d). 1.665 e). 1.678 t with 78 – 1 = 77 df, 90% CI: => t* = 1.665

Video Games and GPA 78 students whose roommates brought a videogame to college: average GPA was 2.84, with a sd of 0.669. 132 students whose roommates did not bring a videogame to college: average GPA was 3.105, with a sd of 0.625. For a 90% CI, what is the relevant sample statistic? a). 2.84 b). 3.105 c). -0.265 d). -1.005 e). 1.678 𝑥 1 − 𝑥 2 = 2.84 – 3.105 = -0.265

Video Games and GPA 78 students whose roommates brought a videogame to college: average GPA was 2.84, with a sd of 0.669. 132 students whose roommates did not bring a videogame to college: average GPA was 3.105, with a sd of 0.625. For a 90% CI, what is the standard error? a). 1.960 b). 0.625 c). -0.265 d). 0.044 e). 0.093 SE = 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 = 0.669 2 78 + 0.625 2 132 =0.093

Videogames and GPA 78 ≥ 30, 132 ≥ 30 t with 78 – 1 = 77 df, 90% CI: 78 ≥ 30, 132 ≥ 30 1. Check conditions: 2. Find t*: t with 78 – 1 = 77 df, 90% CI: => t* = 1.665 3. Calculate statistic: 𝑥 1 − 𝑥 2 = 2.84 – 3.105 = – 0.265 SE = 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 = 0.669 2 78 + 0.625 2 132 =0.093 4. Calculate standard error: 5. Calculate CI: 𝑥 1 − 𝑥 2 ± 𝑡 ∗ ∙𝑆𝐸 =−0.265±1.665∙0.093 = −0.265 ± 0.155 (-0.42, -0.11) 5. Interpret in context: We are 90% confident that getting assigned a roommate who brings a videogame to college decreases a student’s first semester GPA by 0.11 to 0.42 points, on average.

Hypothesis Test for a Difference in Means Section 6.4-HT Hypothesis Test for a Difference in Means

The Pygmalion Effect Pygmalion Effect: the greater the expectation placed upon people, the better they perform Teachers were told that certain children (chosen randomly) were expected to be “growth spurters,” based on the Harvard Test of Inflected Acquisition (a test that didn’t actually exist). The response variable is change in IQ over the course of one year. Source: Rosenthal, R. and Jacobsen, L. (1968). “Pygmalion in the Classroom: Teacher Expectation and Pupils’ Intellectual Development.” Holt, Rinehart and Winston, Inc.

The Pygmalion Effect (a) 2.096 (b) 3.80 (c) 1.813 (d) 0.02 (e) 64 n 𝒙 s Control Students 255 8.42 12.0 “Growth Spurters” 65 12.22 13.3 Does this provide evidence that merely expecting a child to do better actually causes the child to do better? Find the test statistic: (a) 2.096 (b) 3.80 (c) 1.813 (d) 0.02 (e) 64 t= 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 −𝑛𝑢𝑙𝑙 𝑆𝐸 = 𝑥 1 − 𝑥 2 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 = 12.22−8.42 13.3 2 65 + 12.0 2 255 = 3.8 1.813 =2.096

The Pygmalion Effect (a) 2.096 (b) 3.80 (c) 0.98 (d) 0.02 (e) 0.04 n 𝒙 s Control Students 255 8.42 12.0 “Growth Spurters” 65 12.22 13.3 Does this provide evidence that merely expecting a child to do better actually causes the child to do better? Find the p-value when the t-statistic is 2.096: (a) 2.096 (b) 3.80 (c) 0.98 (d) 0.02 (e) 0.04 t with 65 – 1 = 64 df, upper tail p-value = 0.02

The Pygmalion Effect n 𝒙 s Control Students 255 8.42 12.0 “Growth Spurters” 65 12.22 13.3 Does this provide evidence that the Pygmalion Effect exists? (that merely expecting a child to do better actually causes the child to do better?) (a) Yes (b) No

Pygmalion Effect H0: 1 = 2 Ha: 1 > 2 1 = average change in IQ for “growth spurters” 1 = average change in IQ for control 1. State hypotheses: n1 = 65 ≥ 30, n2 = 255 ≥ 30 2. Check conditions: 𝑥 1 − 𝑥 2 =12.22 – 8.42 = 3.8 3. Calculate statistic: SE = 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 = 13.3 2 65 + 12.0 2 255 =1.813 4. Calculate standard error: t= 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 −𝑛𝑢𝑙𝑙 𝑆𝐸 = 3.8 1.813 =2.096 5. Calculate test statistic: 6. Compute p-value: t with 65 – 1 = 64 df, upper tail p-value = 0.02 7. Interpret in context: We have evidence that positive teacher expectations significantly increase IQ scores, on average, in elementary school children.