Question

In a study about the hourly wage (X, in rands) paid to young people aged 14-18 with a holiday job,
the following summarised results were measured for a random sample of 400 young people.
X
400
iD1 xi D 3 296 X
400
iD1 x2i D 27 833:35
Two years ago, the mean hourly wage paid to young people with a holiday job was R8:05, with
standard deviation R1:25. The question arises whether the current hourly wage  is greater than
R8:05. In the questions below, assume that the variation in the hourly wages has not changed.Calculate the mean and standard deviation of the data.
(b) Use a 95% confidence interval to answer the question.
(c) Perform a hypothesis test at the 5% level of significance to answer the question (Make sure
you indicate all steps).
(d) Do you need to know whether X is normally distributed? Why/why not?

Answer 1

Given that

$$\sum_{\mathclap{1\le i\le 400}} X_i=3296, \ \ \ \ \sum_{\mathclap{1\le i\le 400}} X_i^2=27833.35$$

a) calculate the mean and standard deviation of the data

$$\mu=E(X)={1 \over n}\sum_{\mathclap{1\le i\le n}} X_i={1 \over 400}\sum_{\mathclap{1\le i\le 400}} X_i=$$

$$={3296 \over 400}=8.24$$

$$Var(X)=\sigma^2=E(X^2)-(E(X))^2=$$

$$={27833.35 \over 400}-(8.24)^2=1.685775$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{1.685775}\approx1.2984$$

b)We need to construct the $95\%$ confidence interval for the population mean $\mu$ with unknown population variance.

The critical value for $\alpha=0.05$ and $df=n-1=399$ degrees of freedom is

$$t_c=t_{1-\alpha/2;n}=1.966$$

The corresponding confidence interval is computed as shown below:

$$CI=\big(\bar{X}-t_c\times{s \over \sqrt{n}},\ \bar{X}+t_c\times{s \over \sqrt{n}}\big)=$$

$$=\big(8.24-1.966\times{1.2984 \over \sqrt{400}},\ 8.24+1.966\times{1.2984 \over \sqrt{400}}\big)=$$

$$=(8.112,\ 8.368)$$

If we use z-test.

The critical value for $\alpha=0.05$ is $z_c=z_{1-\alpha/2}=1.96.$

The corresponding confidence interval is computed as shown below:

$$CI=\big(\bar{X}-z_c\times{\sigma \over \sqrt{n}},\ \bar{X}+z_c\times{\sigma \over \sqrt{n}}\big)=$$

$$=\big(8.24-1.96\times{1.2984 \over \sqrt{400}},\ 8.24+1.96\times{1.2984 \over \sqrt{400}}\big)=$$

$$=(8.113,\ 8.367)$$

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

$H_0:\mu\leq8.02$

$H_1:\mu>8.02$

This corresponds to a right-tailed test, for which a z-test for one mean, with known population standard deviation $\sigma=1.25$ will be used.

Based on the information provided, the significance level is $\alpha=0.05,$ and the critical value for a right-tailed test is $z_c=1.64.$

The rejection region for this right-tailed test is $R=\{z:z>1.64\}$

The z-statistic is computed as follows:

$$z={\bar{X}-\mu \over \sigma/\sqrt{n}}={8.24-8.05 \over 1.25/\sqrt{400}}=3.04$$

Since it is observed that $z=3.04>1.64=z_c,$ it is then concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean $\mu$ is greater than $8.05,$ at the $0.05$ significance level.

Using the P-value approach: The p-value for $z=3.04$ is $p=0.001183,$ and since $p=0.001182<0.05,$ it is concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean $\mu$ is greater than $8.05,$ at the $0.05$ significance level.

The 95% confidence interval is

$$CI=\big(\bar{X}-z_c\times{\sigma \over \sqrt{n}},\ \bar{X}+z_c\times{\sigma \over \sqrt{n}}\big)=$$

$$=\big(8.24-1.96\times{1.25 \over \sqrt{400}},\ 8.24+1.96\times{1.25 \over \sqrt{400}}\big)=$$

$$=(8.1175,\ 8.3625)$$

d) When the sample size is large, the $z-$tests are easily modified to yield valid test procedures without requiring either a normal population distribution or known standard deviation.