Question #348796

a researcher reports that the average salary of a teacher php 27000 a sample of 32 teachers has a mean of php 29500 at a=0.01 test the claim that the co teachers earn more than php 27000 a months the standard deviation of the population is php 2 250

1
Expert's answer
2022-06-08T13:18:58-0400

The following null and alternative hypotheses need to be tested:

H0:μ27000H_0:\mu\le27000

H1:μ>27000H_1:\mu>27000

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

Based on the information provided, the significance level is α=0.01,\alpha = 0.01, and the critical value for a right-tailed test is zc=2.3263.z_c = 2.3263.

The rejection region for this right-tailed test is R={z:z>2.3263}.R = \{z:z>2.3263\}.

The z-statistic is computed as follows:


z=xˉμσ/n=29500270002250/32=6.2854z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{29500-27000}{2250/\sqrt{32}}=6.2854

Since it is observed that z=6.2854>2.3263=zc,z=6.2854>2.3263=z_c, it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is p=P(z>6.2854)=0,p=P(z>6.2854)= 0, and since p=0<0.01=α,p= 0<0.01=\alpha, it is concluded that the null hypothesis is rejected.

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

is greater than 27000, at the α=0.01\alpha = 0.01 significance level.


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