The manager of a university bank wants to add a new fee to student checking acco

Question

The manager of a university bank wants to add a new fee to student checking accounts. She has determined this is only cost effective if the mean daily balance on these accounts is $150. A random sample of 400 accounts is drawn, for which the sample mean is $142. The manager knows that the accounts are normally distributed with a standard deviation of $60. Based on the information above and a 5% significance level what is the value of the sample mean that is just small enough to reject the null hypothesis (what is the upper limit of the rejection region)? Based on the information above and a significance level of 5%, and the rejection region calculated above, what should the manager do? What is the probability of observing a test statistic at least as extreme as the one computed given that the null hypothesis is true (p-value)?

Explanation / Answer

here std error of mean =std deviation/(n)1/2 =60/(400)1/2 =3

for this is left tailed test; at 0.05 level ;critical value of z =-1.6449

hence upper limit of the rejection region =mean+z*Std error =150-1.6449*3=145.0654

as 142 is below then upper limit of the rejection region ; manager should not  add a new fee to student checking accounts.

test statistic =z=(X-mean)/std error =(142-150)/3 =-2.6667

for above test stat ; p value =0.0038

please revert for any clarification required.,

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