Question

a call centre claims that the employees monthly salary is $ 2400.. you take a sample of 25 employees and find the mean is $2340 and standard deviation is $40 is there enough evidence to reject the director's claim at ~ = 0.05 ? hint t : 0.05 / 2 = 2.06

Answer 1

The sample mean is and the sample standard deviation is s = 40, and the sample size is n=25.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ = 2400

Ha: μ ≠ 2400

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, will be used.

(2) Rejection Region

The significance level is α=0.05, and the critical value for a two-tailed test is

The rejection region for this two-tailed test is

(3) Test Statistics

The t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that ∣t∣ =-7.5 > tc ​= 2.064, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0, and since p = 0 < 0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is different than 2400, at the 0.05 significance level.