Answer:
Check the explanation below.
Step-by-step explanation:
Hello!
The study variable is X: one-time online fundraising that a nonprofit organization has received.
Sample:
n= 50
sample mean X[bar]= $62
Standard deviation S= $9
You are interested in testing the average one-time donation is different than $60. Symbolically: μ ≠ 60
The hypothesis is:
H₀: μ = 60
H₁: μ ≠ 60
You didn't stat any significance level, I'll use a level of 5%
The critical region and the p-value for this test are two-tailed. Since all the available information is from the sample, assuming that the variable has normal distribution I've chosen the Student-t statistic:
t= <u> X[bar] - μ </u>~ t
S/√n
t= <u> 62 - 60 </u>= 1.57
9/√50
The p-value for t=1.57 is 0.1228
Since the p-value: 0.1228 is greater than the significance level of 5%, the decision is to reject the null hypothesis. This means that the average one-time gift online fundraising that a nonprofit organization received is equal to $60.
I hope it helps!