Download presentation
Presentation is loading. Please wait.
Published byMorgan Newton Modified over 6 years ago
1
WARM – UP In a study of heart surgery, one issue was the effects of drugs called beta blockers on the pulse rate of patients during surgery. Subjects were randomly divided into two groups of 30 patients each. One group received a beta blocker, the other a placebo. Pulse rates were recorded during the operations. The treatment group had a mean of 65.2 beats/min. (s = 7.8), while the control group was 70.3 beats/min. (s = 8.3). Give a 95% confidence Interval for the mean difference. Treatment: Control:
2
Treatment: 65.2 7.8 30 Control: 70.3 8.3 30 TWO Sample t – Conf. Int.
95% confidence Interval for the mean difference. TWO Sample t – Conf. Int. SRS – Stated Approximately Normal Distribution – YES because both sample sizes are ≥ 30 [Central Limit Theorem] We can be 95% confident that the true difference in the mean number of beats/min is between and (β Blockers – Placebo) .
3
CH. 24 - The Two Sample t – Test
The Hypothesis: The Mechanics: The t Test Statistic = H0: μ1 = μ2 Ha: μ1 < , >, or ≠ μ2 The P-Value = Ha: μ1 < μ tcdf(-E99, t, df) Ha: μ1 > μ tcdf(t, E99, df) Ha: μ1 ≠ μ (tcdf(|t|, E99, df)) The Conservative Degree of freedom: df = n(smallest) – 1
4
In a study of heart surgery, one issue was the effects of drugs called beta blockers on the pulse rate of patients during surgery. Subjects were randomly divided into two groups of 30 patients each. One group received a beta blocker the other a placebo. Pulse rates were recorded during the operations. The treatment group had a mean of 65.2 beats/min. (s = 7.8), while the control group was 70.3 beats/min. (s = 8.3). Is there evidence that beta blockers reduce pulse rate? μ1 = The true mean number of beats/ min for the treatment group. μ2 = The true mean number of beats/ min for the control group. TWO Sample t – Test SRS – Stated Approximately Normal Distribution – YES because both n = 30 [Central Limit Theorem] H0: μ1 = μ2 Ha: μ1 < μ2 Since the P-Value is less than α = we REJECT H0 . There is evidence that Beta blockers do reduce pulse.
5
Does Buffalo, New York, receive significantly more snow than Chicago, Illinois? To test this question, researchers randomly collected data of yearly snowfall amounts in each city. What conclusions can be made from this data? α = 0.05 Buffalo: 78” 130” 140” 120” 108” 120” 156” 101” Chicago: 84” 96” 114” 81” 98” 103” μ1 = The true mean amount of annual Snow fall in Buffalo, NY. μ2 = The true mean amount of annual Snow fall in Chicago, IL. TWO Sample t – Test SRS – Both Stated Approximately Normal Distribution – Graph BOTH! H0: μ1 = μ2 Ha: μ1 > μ2 Since the P-Value is less than α = the data IS significant (REJECT H0 ). There is evidence that Buffalo’s yearly snowfall average is greater than that of Chicago, IL
6
HW - PAGE 566: 4, 9, 10, 12 H0: μ1 = μ2 Ha: μ1 ≠ μ2 μ true mean test scores μ1 = CPMP μ2 = Trad. SRS Stated Appr Norm: Both n >30 There is a prob. that the statistics will show a difference in means scores, given that there is NO difference. Since the p-val < .05 we Reject H. There is evidence that the two groups are significantly different.
7
HW - PAGE 566: 4, 9, 10, 12
10
Is there evidence to suggest that Mr
Is there evidence to suggest that Mr. Belin’s Period 3 and Period 4 AP Statistics Classes performed differently on Quiz #1? To test this an SRS of students were selected from each class. Period 3: Period 4: μi = The true mean Quiz score of Mr. Belin AP Statistic class (μ3=Period 3 and μ4=Period 4). TWO Sample t – Test SRS – Stated Approximately Normal Distribution – Graph BOTH! H0: μ3 = μ4 Ha: μ3 ≠ μ4 Since the P-Value is NOT less than α = we will Fail to REJECT H0 . There is NO evidence that Period 3 and Period 4 perform differently on quiz #1. NOT real Graph
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.