Question

 An MP claims that the average number of acres in his province’s State Parks is less than 2000 acres. A random sample of five parks is selected and the number of acres is shown below: 959, 1187, 493, 6249, 541 Assume the variable must be normally distributed, at  = 0.01 is there enough evidence to support the claim? To draw your conclusion, state the hypotheses and identify the claim, find the critical value(s), label the acceptance and rejection region, calculate the test value and summarize the results

Answer 1

$$\bar{x}=\dfrac{\displaystyle\sum_{i=1}^nx_i}{n}$$

$$=\dfrac{959+1187+493+6249+54}{5}=1885.8$$

$$s^2=\dfrac{\displaystyle\sum_{i=1}^n(x_i-\bar{x})^2}{n-1}$$

$$=\dfrac{1}{5-1}((959-1885.8)^2+(1187-1885.8)^2+$$

$$+(493-1885.8)^2+(6249-1885.8)^2$$

$$+(541-1885.8)^2)= 6033293.2$$

$$s=\sqrt{s^2}=\sqrt{ 6033293.2}\approx2456.3$$

The following null and alternative hypotheses need to be tested:

$H_0:\mu\geq2000$

$H_1:\mu<2000$

This corresponds to a left-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 $\alpha=0.01,$  

$df=n-1=5-1=4$ degrees of freedom, and the critical value for a left-tailed test is $t_c=-3.746288.$

The rejection region for this left-tailed test is $R=\{t:t<-3.746288\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{1885.8-2000}{2456.3/\sqrt{5}}\approx-0.10396$$

Since it is observed that $t=-0.10396>-3.746288=t_c,$ 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 less than 2000, at the $\alpha=0.01$ significance level.

Using the P-value approach: The p-value for left-tailed, $\alpha=0.01, df=4,$

$t=-0.10396$ is $p=0.461103,$ and since $p=0.461103>0.01=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is less than 2000, at the $\alpha=0.01$ significance level.

Therefore, there is not enough evidence to claim that the average number of acres in province’s State Parks is less than 2000 acres, at the $\alpha=0.01$ significance level.