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The solution to an assignment using minitab software for statistical analysis. The assignment involves conducting t-tests and calculating confidence intervals for the differences between means in two groups. The data presented relates to the mean time children take to answer questions in an 'active' and 'passive' learning style, and the mean time students take to report cheating. The analysis concludes that there is a significant difference in mean time between the 'active' and 'passive' groups, but insufficient evidence to support a difference in mean time between the treatment and control groups.

Typology: Quizzes

2009/2010

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Download Minitab Assignment: Analyzing Data with T-Tests and Confidence Intervals - Prof. Susan Sur and more Quizzes Statistics in PDF only on Docsity! MINITAB Assignment 6 1. a. ————— 11/21/2009 10:03:20 PM ———————————————————— Welcome to Minitab, press F1 for help. ————— 12/3/2010 11:34:39 AM ———————————————————— Welcome to Minitab, press F1 for help. Retrieving project from file: '\\PSF\HOME\DOWNLOADS\MINITAB_6_DATA.MPJ' b.
Probability Plot of the Distribution of Correct Answers for 22 Children Assigned to the "Active" Style Group
Normal - 95% CI
2
Mean 23.05
SiDey 4.248
5 N 22
AD 0.353
0 P-Wslue 0.433
20
70
2
= 60
Do 50
£
D 40
a
30
20
10
5
10 15 20 25 30 35] 40
Number of Correct Answers
Boxplot of the Distribution of Correct Answers for 22 Children Assigned to the "Passive" Style Group
le 14 16 18 20 22 24 26
Number of Correct Answers
vi. At a level of significance of 0.01, there is sufficient evidence to support the claim that the mean level of correct identifications for the children in the “active” style group is greater than the mean level of correct identifications for the children in the “passive” style group. vii. Yes, I would have committed a Type I error, which is when the null hypothesis is rejected while, in reality, it is true. 2. a.
Probability Plot of Distribution of the Time Used Before Subjects Asked for Help in the Treatment Group
Normal - 95% CI
Time (in Seconds)
9
Mean 316.4
SiDev 1367
5 N B
AD Oars
eu P-Wslue 0,198
a0
70
”
& 60
Do 50
i
a 40
a
30
20
10
5
1 f T i T
Q 250 S00 70
Time (in Seconds)
Boxplot of Distribution of the Time Used Before Subjects Asked for Help in the Control Group
100 150 200 250 300 350 400
From the boxplots and probability plots displayed above, we can see that using t procedures would be reasonable with these sets of data. The probability plots show a roughly linear pattern in both of the distributions, with no observations falling outside of the given range. Also, the boxplots show that there is no skewness present or any outliers visible that would have an adverse effect on my analysis. b. MTB > twosample c4 c5; SUBC> confi 90. Two-Sample T-Test and CI: Treatment, Control Two-sample T for Treatment vs Control SE N Mean StDev Mean Treatment 13 316 137 38 Control 13 219 100 28 Difference = mu (Treatment) - mu (Control) Estimate for difference: 97.0 90% CI for difference: (16.2, 177.8) T-Test of difference = 0 (vs not =): T-Value = 2.06 P-Value = 0.051 DF = 22 d. MTB > pone 172 19; SUBC> test .16; SUBC> altern -1; SUBC> usez. Test and CI for One Proportion Test of p = 0.16 vs p < 0.16 95% Upper Sample X N Sample p Bound Z-Value P-Value 1 19 172 0.110465 0.149780 -1.77 0.038 Using the normal approximation. i. H0 : p=0.16Ha : p<0.16 ii. The significance level for this problem is .05. iii. The test statistic for this problem is -1.77. iv. The P-value for this problem is 0.038. v. Since the P-value = 0.038 is less than α = 0.01, or 99%. = 0.05, I reject the null hypothesis. vi. At the 0.05 level of significance, there is sufficient evidence to support the claim that the proportion of all undergraduates at this university who would report cheating has decreased from 0.16. vii. If the true proportion was 0.15, then no error would have been committed, as this would have been valid under the rejection of the null hypothesis.