****Does the correlation .34 between extra curricular activities and academic achievement represent a true relationship? The null hypothesis is there is no relationship betwee extra-curricular activities and academic achievement.
1. HO:=0.00
H1:> 0.00
2. Distribution: t; n=28
3. a: .05
4. T.S.:t=1.843
5. df :27; Significance level:.043
****What is the probability that the r of .34 occurred by chance?
erhd
To determine the probability that the correlation coefficient of 0.34 occurred by chance, we can use the t-distribution and the significance level.
1. First, let's set up the null and alternative hypotheses:
- Null hypothesis (HO): There is no relationship between extra-curricular activities and academic achievement (correlation coefficient = 0).
- Alternative hypothesis (H1): There is a positive relationship between extra-curricular activities and academic achievement (correlation coefficient > 0).
2. Next, we need to determine the distribution to use and the degrees of freedom (df):
- Since we are testing the relationship between two variables, we can use the t-distribution.
- The degrees of freedom (df) is given as 27, which is n - 2, where n is the number of data points used to calculate the correlation coefficient (in this case, n = 28).
3. Now, we need to determine the significance level (a) and the critical t-value:
- The significance level (a) is given as 0.05, which means we are willing to accept a 5% chance of making a Type I error (rejecting the null hypothesis when it is true).
- From the t-distribution table or calculator, find the critical t-value for df = 27 and a one-tailed test with a = 0.05. In this case, the critical t-value is 1.706.
4. Calculate the test statistic (t):
- The test statistic (t) is calculated as the correlation coefficient divided by the standard error of the correlation coefficient. In this case, t = 1.843.
5. Determine the probability (p-value):
- Finally, we can determine the probability (p-value) associated with the test statistic. This is the probability of obtaining a t-value as extreme or more extreme than the observed test statistic, assuming the null hypothesis is true.
- Using a t-distribution calculator with df = 27 and a one-tailed test, find the probability associated with a t-value of 1.843. In this case, the p-value is approximately 0.043.
Therefore, the probability that the correlation coefficient of 0.34 occurred by chance (assuming the null hypothesis is true) is approximately 0.043.