Question

weekly wages of 10 workers at random from factory are given below

Answer 1

We have sample values 578,572,570,568,572,578,570,572,596,584, sample size n=10.

Sample mean

$$\bar{x}=\dfrac{1}{10}(578+572+570+568+572$$

  

$$+578+570+572+596+584)=576$$

Sample variance

$$s^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n-1}=\dfrac{1}{10-1}(4+16+36+64$$

$$+16+4+36+16+400+64)=\dfrac{656}{9}$$ $$s=\sqrt{s^2}=\sqrt{\dfrac{656}{9}}=\dfrac{4\sqrt{41}}{3}\approx8.5375$$

The following null and alternative hypotheses need to be tested:

$H_0:\mu=580$

$H_1:\mu\not=580$

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

Based on the information provided, the significance level is $\alpha = 0.05,$ $df=n-1=9$ and the critical value for two-tailed test is $t_c =2.262156.$

The rejection region for this two-tailed test is $R = \{t:|t|>2.262156\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{576-580}{8.5375/\sqrt{10}}=-1.4816$$

Since it is observed that $|t|=1.4816<2.262156=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for two-tailed, $df=9$ degrees of freedom, $t=-1.4816$ is $p=0.172588,$ and since $p=0.172588>0.05=\alpha,$ 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 580, at the $\alpha = 0.05$ significance level.