In: Statistics and Probability
In a study of parents who have children with ADHD, a simple random sample of parents were asked to rate their overall stress level. A higher score indicates greater stress. The mean rating for the 50 parents in the study was reported to be 46.8 with a standard deviation of 10.34. The researchers considered a score greater than 45 to represent extreme stress. Is there evidence that the mean stress level for the parents in this study is above this level? Conduct a hypothesis test at the 5% significance level by completing the problems below.
A. Write the hypotheses for this test.
B. Write two sentences checking whether the assumptions of the test have been met.
C. What is the probability of making a type I error? Describe a type I error in terms of the problem.
D. What is the t-crit value for this test? Round your answer to 2 decimal places.
E. Write the decision rule for the test.
F. Compute the value of t-calc. Round your answer to two decimal places.
G. Based on your value of t-calc, what decision should you make? Reject or fail to reject Ho
H. Write a conclusion that anyone could understand. Make sure to give the significance level of the test
Ho : Mean score equal to 45
Ha : Mean score greater than 45
Ho : µ = 45
Ha : µ > 45
(Right tail test)
a)Random sample
b) n=50 means sample is large enough to
normal
Probability of type 1 error = Level of Significance ,
α = 0.05
sample std dev , s = 10.3400
Sample Size , n = 50
Sample Mean, x̅ = 46.8000
degree of freedom= DF=n-1= 49
Standard Error , SE = s/√n = 10.3400 / √
50 = 1.4623
critical t value, t* =
1.6766 [Excel formula =t.inv(α/no. of tails,df)
]
t-test statistic= (x̅ - µ )/SE = (
46.800 - 45 ) /
1.4623 =
1.23
t critical > t calc , Do not reject null
hypothesis
There is not sufficient evidence that Mean score greater than 45.
Please revert back in case of any doubt.
Please upvote. Thanks in advance.