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Basic Question 1 of 10

A consumer group wants to prove that average hospital costs are more than \$931 per day. The group randomly samples 60 accounts and finds a sample mean of \$950. The hypothesis test is at an α-level of 5% and assumes a σ of \$50. The critical value is ______ standard deviations.

A. 1.645
B. -1.645
C. 1.28

User Comment
Bibhu One can refer to www.mathsnet.net/asa2/modules/s33binom.html for further reading on this subject. The easier way is sample mean 950 above 931. So +Z and then 5% means 1.645.
whiteknight we can approach the problem like this...for 5% level of significance in a one tail test the z value is 1.645...i.e the right hand portion of the normal distribution....thus for any Z greater than 1.645 this statement will hold true ...
raymondg could somebodyplease explain how to arrive to 1.645 ?
meghanchloe I THOUGHT 5% LEVEL OF SIGNIFICANT IS 1.96
ebayer 5% one tail = 10% both tails -> use 1.645 instead of 1.96.
johntan1979 This test is clearly one-tailed (>931). So answer should be 1.645 if alpha is .05. two-tailed would be 1.96.
Sp1993 Hi johntan1979, could you please clarify why a two-tailed test would give an answer of 1.96! Thanks!