Answer to Question #349356 in Statistics and Probability for Cecel

Question

It is claimed that the mean annual salary of call center customer service representative is php 188 584.00. A researcher randomly selected 45 call center customer service representative. He computed the mean of their annual salaries and obtained a mean of php 188 600.00. Does this show that the mean salary of call center customer service representatives is greater than php 188 584.00? Use 0.05 level of significance and assume that the population standard deviation is php 39.50.

Answer 1

Solution:

$X=188600$ php. - sample mean;

$\mu=188584$ php. - population mean;

$\sigma=39.5$ - standard deviation;

$n=45$ - number of samples;

$\alpha=0.05$ - level of significance;

Ho - null hypothesis: $\mu<=188584;$

Ha- alternative hypothesis: $\mu>188584;$

Let's use test statistics.

$z=\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}=\frac{188600-188584}{\frac{39.5}{\sqrt{45}}}=2.7172;$

z- score of $z=2.7172$ equal 0.4966+05;

It means 99.66% salary less than 188600 and less than 1% greater than 188600.

So, if we use level of significance:

$1-\alpha=1-0.05=0.95;$ $z_c= 1.96$ then $z_c<z;$

If the absolute value of the z-value is greater than the critical value, means you reject the null hypothesis.

Answer:

The result: mean salary of representatives greater then 188584 php.