Question

Assume that z is the test statistic.

(a) H₀: μ = 22.5, H_a: μ > 22.5; x = 25.9, σ = 7.4, n = 33
(i) Calculate the test statistic z. (Round your answer to two decimal places.)


(ii) Calculate the p-value. (Round your answer to four decimal places.)


(b) H₀: μ = 200, H_a: μ < 200; x = 193.8, σ = 35, n = 36
(i) Calculate the test statistic z. (Round your answer to two decimal places.)


(ii) Calculate the p-value. (Round your answer to four decimal places.)


(c) H₀: μ = 12.4, H_a: μ ≠ 12.4; x = 11.3, σ = 5.3, n = 26
(i) Calculate the test statistic z. (Round your answer to two decimal places.)


(ii) Calculate the p-value. (Round your answer to four decimal places.)

Answer 1

(a) H₀: μ = 22.5, H_a: μ > 22.5;    This is right tailed test

Given that, x = 25.9,        σ = 7.4,            n = 33

(i) Therefore, the test statistic for the problem will be,

$z=\frac{x-\mu}{\sigma/\sqrt{n}} = \frac{25.9-22.5}{7.4/\sqrt{33}}=\frac{3.4}{1.2882}=2.6394$
(ii) The p-value for the right tailed test corresponds to probability $P(Z > z)$ which can be obtained using Excel function NORMSDIST(z) as below:
$p-value=P(Z > z)=1-P(Z\leqz) =1-NORMSDIST(2.6394)=(1-0.9958)=0.0042$

(b) H₀: μ = 200, H_a: μ < 200;    This is left tailed test

Given that, x = 193.8,        σ = 35,            n = 36

(i) Therefore, the test statistic for the problem will be,

$z=\frac{x-\mu}{\sigma/\sqrt{n}} = \frac{193.8-200}{35/\sqrt{36}}=\frac{-6.2}{5.8333}=-1.0629$

(ii) The p-value for the left tailed test corresponds to probability $P(Z \leq z)$ which can be obtained using Excel function NORMSDIST(z) as below:
$p-value=P(Z\leq z) =NORMSDIST(-1.0629)=0.1439$

(c) H₀: μ = 12.4, H_a: μ ≠ 12.4    This is left tailed test

Given that, x = 11.3,        σ = 5.3,            n = 26

(i) Therefore, the test statistic for the problem will be,

$z=\frac{x-\mu}{\sigma/\sqrt{n}} = \frac{11.3-12.4}{5.3/\sqrt{26}}=\frac{-1.1}{1.0394}=-1.0583$

(ii) The p-value for the two tailed test corresponds to probability $2P(Z \leq z)$ which can be obtained using Excel function NORMSDIST(z) as below:
$p-value=2P(Z\leq z) =2NORMSDIST(-1.0583)=2(0.1450)=0.2899$