# :≥10 :<10 A sample of 50 provided a sample mean of 9.49 and sample standard deviation of 2. 1. At α=.05, what is the rejection rule?Find critical value for z

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We are given `n=50,bar(x)=9.49,s=2,alpha=.05` (And it appears that you wanted `mu=10` ) You are asking for the critical value.

** You provided a sample mean -- if you only know the sample mean you will not use a z-test; you would use a t-test with the student's t-table. I am proceeding under the assumption that you meant that we knew the population standard deviation, or `sigma=2` . The critical values would be affected if this really is a sample standard deviation.)**

The critical value is affected by the sample size, the mean and standard deviation, the significance level, and by the type of test (one tailed or two tailed).

(1) Suppose the null hypothesis is `H_0:` `mu=10` and the alternative hypothesis is `H_1:mu != 10` . This is a two tailed test and we would find `z_(alpha/2)` for the critical values.

Now `z_(.025)` gives critical values of `+-1.96` from the standard normal table. Thus the critical region would be `z<-1.96,z>1.96`

The test value would be `z=(9.49-10)/(2/sqrt(50))=-1.803` which is not in the critical region so you would not reject the null hypothesis.

(2) Suppose `H_0: mu=10` and `H_1:mu<10` . Then the critical value is given by `z_(.05)=-1.64` and the critical region is z<-1.64.

The test value does not change, but now it is in the critical region so you would reject the null hypothesis and find sufficient evidence to reject the null hypothesis.