Question

Determine the given, formulate the null and alternative hypothesis in words and in

symbols, and the appropriate test statistic.

A company produced ethyl alcohol and claimed to have a mean alcohol content of 70%. A random sample of 80 of ethyl alcohol was take as sample to verify this claim. It was found out that the mean alcohol content is 65% with a standard deviation of 2%. Test the claim at 5% level of significance and assume that the population is normally distributed.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=70,$ alcohol content is 70%

$H_a:\mu\not=70,$ alcohol content is not 70%

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 $\alpha = 0.05,$ $df=n-1=79$ degrees of freedom, and the critical value for a two-tailed test is $t_c = 1.99045.$

The rejection region for this two-tailed test is $R = \{t: |t| > 1.99045\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{65-70}{2/\sqrt{80}}\approx-22.3607$$

5. Since it is observed that $|t| =22.3607>1.99045=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=79$ degrees of freedom, $t=-22.3607$ is $p=0,$ and since $p=0<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is different than 70, at the $\alpha = 0.05$ significance level.