Answer to Question #165527 in Math for ...

Question

A manager claims that the company has given an average salary increment of RM 40

to all the employees per year. A researcher wants to check if this claim is true. A

random sample of 45 employees taken by the researcher showed that these employees

got an average increment of RM 38 per year with a standard deviation of RM 8.

a) Is there enough evidence to accept the manager’s claim at 0.1 level of significance?

b) Test at 2.5% significance level whether the average salary increment of all

employees is less than RM 40 per year.

Answer 1

a) The provided sample mean is $\bar{x}=38$ and the sample standard deviation is $s=8.$

The sample size is $n=45.$ The number of degrees of freedom are $df=n-1=44,$ and the significance level is $\alpha=0.1.$

The following null and alternative hypotheses need to be tested:

$H_0: \mu=40$

$H_1: \mu\not=40$

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

Based on the information provided, the critical value for a two-tailed test is $t_c=1.68023.$

The rejection region for this two-tailed test is $R=\{t:|t|>1.68023\}$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{38-40}{8/\sqrt{45}}=-1.67705$$

Since it is observed that $|t|=1.67705<1.68023=t_c,$ it is then concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than 40, at the 0.1 significance level.

Using the P-value approach: The p-value $df=44, t=-1.67705,$ $two-tailed,$ is $p=0.100634,$ and since $p=0.100634>0.1,$ it is concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than 40, at the 0.1 significance level.

b) The provided sample mean is $\bar{x}=38$ and the sample standard deviation is $s=8.$

The sample size is $n=45.$ The number of degrees of freedom are $df=n-1=44,$ and the significance level is $\alpha=0.025.$

The following null and alternative hypotheses need to be tested:

$H_0: \mu\geq40$

$H_1: \mu<40$

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

Based on the information provided, the critical value for a left-tailed test is $t_c=-2.015368$

The rejection region for this left-tailed test is $R=\{t:t<-2.015368\}$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{38-40}{8/\sqrt{45}}=-1.67705$$

Since it is observed that $t=-1.67705>-2.015368=t_c,$ it is then concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population mean $\mu$ is les than 40, at the 0.025 significance level.

Using the P-value approach: The p-value $df=44, t=-1.67705,$ $left-tailed,$ is $p=0.0503,$ and since $p=0.0503>0.025,$ it is concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population mean $\mu$ is less than 40, at the 0.025 significance level.