Answer to Question #106501 in Statistics and Probability for Palesa

Question

in a study about the hourly wage 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. Ê400^Xi=3296 Ê400^Xi2^= 27833.35. a) calculate the mean and standard deviation of the data. b) use 95% confidence interval to answer the question. c) perform a hypothesis test at the 5% level of significance to answer the question. d) do you need to know why X is normally distributed?

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

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

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

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

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

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

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:

$H_0: \mu=8$

$H_1: \mu\not=8$

This corresponds to a 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 significance level is $\alpha=0.05,$ and the critical value for a two-tailed test is $t_c=1.966.$  

The rejection region for this two-tailed test is $R=\{t:|t|>1.966\}.$  

The t-statistic is computed as follows:

$$t={\bar{X}-\mu_0 \over s/\sqrt{n}}={8.24-8 \over 1.2984/\sqrt{400}}\approx3.697$$

Since it is observed that $|t|=3.697>1.966=t_c,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is different than $8,$ at the $0.05$ significance level.

Using the P-value approach: The p-value for $t=3.697, \alpha=0.05$ is $p= 0.000249.$ Since $p= 0.000249<0.05,$ it is concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean $\mu$ is different than $8,$ at the $0.05$ significance level.

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.