Answer to Question #349327 in Statistics and Probability for anne

Question

The manager of Penawar Chemical Company claims that their major product contains on average u = 3.0 fluid Ounces of acid material per gallon. If there is too much acid material, the product will be too dangerous. A quality inspector was brought in to test the product. She randomly selected a sample of 100 gallon-size containers of the product. The results were as follow. Mean weight Frequency 2.80 10 2.90 6 3.00 30 3.10 54 SUM 100 d. Using a 5% significance level, explore the critical point. Hence, write the decision rule. [2 marks) e. At the 5% level of significance, investigate whether there is sufficient evidence to support the Company's claim. Assume the acid materials fluid Ounces per gallon are normally distributed. What is your expectation about the company product? (5 marks)

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c} Mean\ weight, x_i &Frequency, f_i\\ \hline 2.80 & 10\\ \hdashline 2.90 & 6\\ \hdashline 3.00 & 30\\ \hdashline 3.10 & 54\\ \hdashline Sum=& 100\\ \end{array}$$

$$\bar{x}=E(X)=\dfrac{1}{N}\sum _ix_if_i$$

$$=\dfrac{2.8(10)+2.9(6)+3.0(30)+3.1(54)}{100}=3.028$$

$$s^2=\dfrac{1}{N-1}\sum _i(x_i-\bar{x})^2\cdot f_i$$

$$=\dfrac{1}{100-1}((2.8-3.028)^2(10)+(2.9-3.028)^2(6)$$

$$+(3.0-3.028)^2(30)+(3.1-3.028)^2(54))$$

$$=\dfrac{1}{99}=\dfrac{0.1024}{11}$$

$$s=\sqrt{\dfrac{0.1024}{11}}\approx0.0965$$

d. The following null and alternative hypotheses need to be tested:

$H_0:\mu\le3.0$

$H_1:\mu>3.0$

This corresponds to a right-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 $\alpha = 0.05,$ $df=n-1=99$ and the critical value for a right-tailed test is $t_c =1.660391.$

The rejection region for this right-tailed test is $R = \{t:t>1.660391\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{3.028-3.0}{0.0965/\sqrt{100}}=2.9016$$

Since it is observed that $t=2.9016>1.660391=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for right-tailed, $df=99$ degrees of freedom, $t=2.9016$ is $p=0.002288,$ and since $p=0.002288<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

e. Therefore, there is enough evidence to claim that the population mean $\mu$ is greater than 3.0, at the $\alpha = 0.05$ significance level.

Therefore, there is enough evidence to claim that the product will be too dangerous, at the $\alpha = 0.05$ significance level.