Question #341101

The average height of students in a freshman class of a certain school has been 158.55 cm with a population standard deviation of 8.8 cm. Is there a reason to believe that there has been a change in the average height if a random sample of 51 students in the present freshman class has an average height of 154.25 cm? Use a 0.1 level of significance.





What are the given? Write only the number. :



population mean:Blank 1 cm



population standard deviation:Blank 2 cm



sample size:Blank 3



sample mean:Blank 4 cm



level of significance:Blank 5





What are the critical values? Write the positive critical value first then the negative.



z:Blank 6 andBlank 7





What is the value of the calculated z? Round your answer to the nearest hundredths.



z:Blank 8

1
Expert's answer
2022-05-16T16:22:15-0400

population mean: 158.55158.55 cm

population standard deviation: 8.88.8 cm

sample size:5151

sample mean: 154.25154.25 cm

level of significance: 0.10.1

What is the critical value?

zc1=1.6449,zc2=1.6449z_{c1}=-1.6449, z_{c2}=1.6449

What is the value of the calculated z?


z=xˉμσ/n=154.25158.558.8/513.49z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{154.25-158.55}{8.8/\sqrt{51}}\approx-3.49

The following null and alternative hypotheses need to be tested:

H0:μ=158.55H_0:\mu=158.55

H1:μ158.55H_1:\mu\not=158.55

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Since it is observed that z=3.49<1.6449=zc1,z=-3.49<-1.6449= z_{c1}, it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is p=2P(z<3.49)=154.25158.55,p=2P(z<-3.49)=154.25-158.55, and since p=154.25158.55<0.1=α,p=154.25-158.55<0.1=\alpha, it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean μ\mu

is different than 158.55, at the α=0.1\alpha = 0.1 significance level.


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