A sample of 22 observations provides the following statistics: Use Table 2. sx =

Question

A sample of 22 observations provides the following statistics: Use Table 2. sx = 15, sy = 17, and sxy = 119.86 a-1. Calculate the sample correlation coefficient rxy. (Round your answer to 2 decimal places.) Sample correlation coefficient a-2. Interpret the sample correlation coefficient rxy. The correlation coefficient indicates a positive linear relationship. The correlation coefficient indicates a negative linear relationship. The correlation coefficient indicates no linear relationship. b. Specify the hypotheses to determine whether the population correlation coefficient is positive. H0: xy 0; HA: xy < 0 H0: xy 0; HA: xy > 0 H0: xy = 0; HA: xy 0 c-1. Calculate the value of the test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) Test statistic c-2. Approximate the p-value. 0.01 < p-value < 0.025 0.100 < p-value < 0.200 0.050 < p-value < 0.100 p-value > 0.200 d. At the 10% significance level, what is the conclusion to the test? Do not reject H0; we cannot state the population correlation is positive. Reject H0; we cannot state the population correlation is positive. Do not reject H0; we can state the population correlation is positive. Reject H0; we can state the population correlation is positive.

Explanation / Answer

(a-1) rxy=sxy/(sx*sy)=119.86/(15*17)=0.47

(a-2) The correlation coefficient indicates a positive linear relationship.

since the rxy is positive

(b) H0: xy 0; HA: xy > 0

(c1) n=22 and r=0.47

we use t-test and t =r/sqrt[(1—r2)/(n—2)]=0.47/sqrt((1-0.47*0.47)/(22-2))=2.38 with n-2=20 df

(c2) one tailed p-value=0.0137 ( using ms-excel=TDIST(2.38,20,1))

Approximate the p-value. 0.01 < p-value < 0.025

(c3)Reject H0; we can state the population correlation is positive.

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