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Chapter 8: Q. 9.78 (page 380)

We have been provided a sample mean, sample size, and population standard deviation. In the given case, use the one-mean z-test to perform the required hypothesis test at the 5%significance level.

x炉=21,n=32,蟽=4,H0:渭=22,Ha:渭<22

Short Answer

Expert verified

The value of z is -1.41, critical value is-1.645,P=0.002and rejectH0.

Step by step solution

01

Step 1. Given information.

Consider the given question,

x炉=21,n=32,蟽=4,H0:渭=22,Ha:渭<22

02

Step 2. Consider the test hypothesis.

Consider the given hypothesis,

渭is the population mean.

The test hypothesis,

H0:渭=22VsHa:渭<22

Therefore, the test is left tailed test.

And the level of significance is伪=0.05.

03

Step 3. Use the test statistics.

We want to find the hypothesis test about the mean 渭,

z=x炉-渭0蟽n=21-22432=-1.41

Therefore, this is left tailed test with 伪=0.05, the critical value is -za=-z0.05=-1.645

The rejection region is z<-z0.05.

Here, data-custom-editor="chemistry" z=-1.41>-1.645.

Therefore, we do not reject H0at 5% level of significance as the value of the test statistics z does not fall in the rejection region.

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