1. A) The Perceived Stress Scale developed by Cohen (1988) has a mean score of 1
ID: 3309531 • Letter: 1
Question
1. A) The Perceived Stress Scale developed by Cohen (1988) has a mean score of 13.02, and a standard deviation of 7.35, for the general population of adults. A researcher in a college health center is interested in determining if the perceived stress of new mothers students differs from the general population. She surveys a random sample of 16 new mothers who have just given birth to their first child She finds the mean perceived stress score for this sample was 16.4. Test the null hypothesis at the .05 level of significance. Include the step by step hypothesis test summary illustrated in the text on p. 187 and in my class notes. (ATTACHED***)
B) Statistical decisions are based on probability and therefore, they may be correct or in error. Given the decision that you made in Part A of this problem, if your decision is in error, what kind of error would you have made? What can you do to decrease the probability of making this error?
disruplive ity of loyees? Because of our emphasis on hypothesis testing, research problems i car in thisbook as finished products, usually in the first one or two sentences of a new example HYPOTHESIS TEST SUMMARY: z TEST FOR A POPULATION MEAN (SAT SCORES) Research Problem Does the mean SAT math score for all local freshmen differ from the national average of 500? Statistical Hypotheses Ho : = 500 H1:#500 Decision Rule Reject H, at the.05 level of significance if z2 1.96 or if zs-1.96. Calculations Given 110 n 100 533-500 Decision Reject H, at the .05 level of significance because 3 exceeds 1.96. Interpretation The mean SAT math score for all local freshmen does not equal-it exceeds-the national average of 500Explanation / Answer
Sol:
Research problem:
if the perceived stress of new mothers students differ from the general population.
Statistical Hypothesis:
H0:=13.02
H1: 13.02
Decsion rule:
Reject Ho at the 0.05 level of significance if z>=1.96 or if z<=1.96
Calculations:
Given X bar=16.4
hyp=13.02
sigma xbar=Sigma/sqrt(n)
=7.35/sqrt(16)
=1.8375
Z=16.4-13.02/1.8375
=1.839
=1.84
Decsion:
Fail to reject Ho at the 0.05 level of significance because z=1.84 is below 1.96.
Intrpretation:
perceived stress of new mothers students does not differs form the general popualtion.
SolutionB:
We would have made Type1 error.
to decrease the probability of making this error increase alpha(level of significance)
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