Answer to Question #341101 in Statistics and Probability for neeca

Question

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

Answer 1

population mean: $158.55$ cm

population standard deviation: $8.8$ cm

sample size:$51$

sample mean: $154.25$ cm

level of significance: $0.1$

What is the critical value?

$z_{c1}=-1.6449, z_{c2}=1.6449$

What is the value of the calculated z?

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

The following null and alternative hypotheses need to be tested:

$H_0:\mu=158.55$

$H_1:\mu\not=158.55$

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

Using the P-value approach:

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

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

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