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

x=20,n=24,σ=4,H0:μ=22,Ha:μ22

Short Answer

Expert verified

Ans: We have to reject H0at5%the level of significance as the value of z falls in the rejection region.

Step by step solution

01

Step 1. Given information.

given,

x=20,n=24,σ=4,H0:μ=22,Ha:μ22

02

Step 2. Let's perform a hypothesis test.

H0:μ=22Ha:μ22

Where μis the population mean.

Since the alternative hypothesis contains the symbol localid="1651574793371" .

And the test is a two-tailed test.

Now,

The level of significance is, α=0.05.

03

Step 3. Now, we find the hypothesis test about the mean, μ.

we have,

x=20,n=24,σ=4

Test statistic

z=x¯μ0σn=2022424=2.45

04

Step 4. Since this is a two-tailed test with α=0.05,  

By using the normal distribution table, the critical values are:

±zα/2=±z0.05/2=±z0.025=±1.96

Here the rejection regions are z<-zα/2orz>zα/2

i.e., z<-1.96orz>1.96

Here z=-2.45<-z0.025=-1.96

Since the level of significance as the value of z falls in the rejection region.

So, we have to reject localid="1651211012259" H0at5%.

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Most popular questions from this chapter

Define theP- value of the hypothesis test.

The normal probability curve and stem-and-leaf diagram of the data are shown in figure; σis known.

Perform Hypothesis test for mean of the population from which data is obtained and decide whether to use z-test, t-test or neither. Explain your answer.

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(b) Type II error.

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