Answer to Question #350162 in Statistics and Probability for qqqq

Question #350162

a researcher reports that the average salary of college deans is more than Php 63,000 a sample of 35 college deans has mean salary of php 65,700




1
Expert's answer
2022-06-13T14:55:06-0400

The following null and alternative hypotheses need to be tested:

H0:μ63000H_0:\mu\le 63000

H1:μ>63000H_1:\mu>63000

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=65700630005250/35=3.0426z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{65700-63000}{5250/\sqrt{35}}=3.0426

Since it is observed that z=3.0426>2.3263=zc,|z|=3.0426>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>3.0426)=0.001173,p=P(z>3.0426)=0.001173, and since p=0.001173<0.01=α,p=0.001173<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 63000, at the α=0.01\alpha = 0.01 significance level.


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