Question

A researcher claims that adult hogs fed a special diet will have an average weight of 200 pounds. A sample of 10 hogs has an average weight 198.2 pounds and standard deviation of 3.3 pounds. At a= 0.05, can the claim be rejected? Also, find the 95% confidence interval of the true mean?

Answer 1

The critical value for $\alpha = 0.05$ and $df = n-1 = 9$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.262156.$

The corresponding confidence interval is computed as shown below:

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

$$=(198.2-2.262156\times\dfrac{3.3}{\sqrt{10}},$$

$$198.2+2.262156\times\dfrac{3.3}{\sqrt{10}})$$

$$=(195.839,200.561)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $195.839 < \mu < 200.561 ,$ which indicates that we are 95% confident that the true population mean $\mu$  is contained by the interval $(195.839, 200.561).$

The following null and alternative hypotheses need to be tested:

$H_0:\mu=200$

$H_a:\mu\not=200$

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 $\alpha = 0.05,$ $df=n-1=9$ degrees of freedom, and the critical value for a 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:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{198.2-200}{3.3/\sqrt{10}}=-1.7249$$

Since it is observed that $|t| = 1.7249 \le 2.262156=t_c ,$ 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, $t=-1.7249$ is $p=0.118631,$ and since $p=0.118631>0.05=\alpha,$ 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 200, at the $\alpha = 0.05$ significance level.