Answer:
a) The null and alternative hypothesis are:
[tex]H_0: \mu_d=0\\\\H_a:\mu_d> 0[/tex]
b) Test statistic t = 2.857
P-value = .005
c. Yes. The null hypothesi is rejected.
At a significance level of .05, there is enough evidence to support the claim that the effects of priming a participant affect the outcome significantly.
Step-by-step explanation:
This is a paired samples t-test for the population mean difference.
The claim is that the effects of priming a participant affect the outcome significantly.
Then, the null and alternative hypothesis are:
[tex]H_0: \mu_d=0\\\\H_a:\mu_d> 0[/tex]
The significance level is .05.
The sample has a size N=20.
The sample mean difference is MD=8.
The standard error of the mean is SEM=2.8.
Then, we can calculate the t-statistic as:
[tex]t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{8-0}{2.8}=\dfrac{8}{2.8}=2.857[/tex]
The degrees of freedom for this sample size are:
[tex]df=n-1=20-1=19[/tex]
This test is a right-tailed test, with 19 degrees of freedom and t=2.857, so the P-value for this test is calculated as (using a t-table):
[tex]\text{P-value}=P(t>2.857)=0.005[/tex]
As the P-value (.005) is smaller than the significance level (.05), the effect is significant.
The null hypothesis is rejected.
There is enough evidence to support the claim that the effects of priming a participant affect the outcome significantly.