Answer to Question #333991 in Statistics and Probability for Hansi

Question

Assume that the average monthly earnings for the estate workers to be Rs 8500. A sample of 100 garment workers shows a sample mean is Rs 8000. Use a population standard deviation of Rs 1000.

(i) Formulate the null and alternative hypotheses that can be used to determine whether the sample data support the conclusion that the population mean wage is less than that of Rs 8500. (04 marks)

(ii) What is the value of test statistic?

(iii) What is the p-value?

(iv) What was the confidence interval?

(v) At a = 0.01, what is your conclusion. 

Answer 1

(i)The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge8500$

$H_1:\mu<8500$

(ii)

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

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{8000-8500}{1000/\sqrt{100}}=-5$$

(iii)

The p-value is $p =P(Z<-5)= 0.$

(iv)

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-z_c\times\dfrac{\sigma}{\sqrt{n}}, \bar{x}+z_c\times\dfrac{\sigma}{\sqrt{n}})$$

$$=(8000-z_c\times\dfrac{1000}{\sqrt{100}}, 8000+z_c\times\dfrac{1000}{\sqrt{100}})$$

(v)

Based on the information provided, the significance level is $\alpha = 0.01,$ and the critical value for a left-tailed test is $z_c =-2.3263.$

The rejection region for this left-tailed test is $R = \{z: z < -2.3263\}.$

Since it is observed that $z = -5 < -2.3263=z_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: since the p-value is $p = 0<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is less than 8500, at the $\alpha = 0.01$ significance level.

The critical value for $\alpha = 0.01$ is $z_c = z_{1-\alpha/2} = 2.5758.$

The corresponding confidence interval is

$$CI=(8000-2.5758\times\dfrac{1000}{\sqrt{100}},$$

$$8000+2.5758\times\dfrac{1000}{\sqrt{100}})$$

$$=(7742.42,8257.58 )$$