Suppose you obtained a Spearman rs of 0.53 from a sample of 13 pairs of scores. For a two-tailed test, such a correlation coeffi
1 answer:
Answer:
C: p < 0.01
Step-by-step explanation:
We are given;
Spearman rank Correlation Coefficient: rs = 0.53
sample size: n = 13 pairs
Now, from the table attached, tracing n = 13 and locating a corresponding value of rs = 0.53 which falls in between 0.484 and 0.56. Thus, we can see that p is greater than the nominal significance value of 0.05 but less than 0.01.
Thus, correct answer is p < 0.01
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Answer:
Case n =5
Case n =15
Case n = 40
P value
Case n =5
Case n =15
Case n =40
Step-by-step explanation:
Data given and notation
represent the sample mean
represent the population standard deviation
sample size
represent the value that we want to test
represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is lower than 5, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:
(1)
z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
Case n =5
Case n =15
Case n = 40
P-value
Since is a left tailed test the p value would be:
Case n =5
Case n =15
Case n =40