I just really need help. My teacher hasn\'t explained how to do any of this and
ID: 3243903 • Letter: I
Question
I just really need help. My teacher hasn't explained how to do any of this and won't explain it either I tried. Please Help Me!!!!! I am not sure how to answer these questions with the data provided.
Consider a possible linear relationship between two variables that you would like to explore.
1.Define the relationship of interest and a data collection technique.
3.Construct a model of the relationship and evaluate the validity of that model.
4. Describe the techniques and procedures that will be used.
5. State general equipment that will be necessary and its intended use.
6. Describe type of data that will be collected and measurements that will be made.
7. Define and describe experimental units and be specific.
This is the data from the systolic blood pressure;
A random sample of n=30 was taken from each of three populations of young males with 10 in each group. Systolic blood pressure measurements were taken on each child.
´Hypothesis: H0; u1=u2=u3 vs. H1; u1 not= u2 not= u3
´The a-level : a = 0.05
´ The test statistic: ANOVA and Descriptive Statistics
´Reject H0 if F.95(2,27)=3.35
Here are the results from the tests:
Descriptive Statistics 1 2 3 Total: Mean 105.8 Mean 97.2 Mean 96 99.655 Standard Error 2.803965446 Standard Error 1.496662955 Standard Error 3.451247761 1.773 Median 108 Median 97 Median 97 Mode 110 Mode 96 Mode #N/A Standard Deviation 8.86691729 Standard Deviation 4.732863826 Standard Deviation 10.9138037 9.548 Sample Variance 78.62222222 Sample Variance 22.4 Sample Variance 119.1111111 91.163 Kurtosis -0.204018374 Kurtosis 0.312196307 Kurtosis -1.171989864 Skewness -0.3281996 Skewness -0.59613559 Skewness 0.041027052 Range 30 Range 16 Range 32 Minimum 90 Minimum 88 Minimum 80 Maximum 120 Maximum 104 Maximum 112 Sum 1058 Sum 972 Sum 960 Count 10 Count 10 Count 10 n=30 Confidence Level(95.0%) 6.343010518 Confidence Level(95.0%) 3.385686823 Confidence Level(95.0%) 7.807264844 Anova: Single Factor SUMMARY Groups Count Sum Average Variance 1 10 1058 105.8 78.62222222 2 10 972 97.2 22.4 3 10 960 96 119.1111111 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 571.4666667 2 285.7333333 3.894003634 0.032667196 3.354130829 Within Groups 1981.2 27 73.37777778 Total 2552.666667 29Explanation / Answer
Note that p-value is less than alpha
as(0.0326671 < 0.05) observe the lower table (2nd column from right)
hence we reject the null hypothesis
and conclude that there is significant evidence that there is difference in three population means
you could have come to same conclusion by observeing F-value = 3.894 > 3.354 (critical value )
see in the lower table (3rd column from right)
we reject the null .