Answer to Question #278267 in Statistics and Probability for nick

Question

Ahmad Fishing Line Inc. produces 10kg test line. 12 randomly selected spools are subjected to tensile-strength tests. The results are as follows:

9.8 10.2 9.8 9.8 9.6 9.1 9.7 9.7 10.1 9.4 10.1 9.7

Use the fact that tensile strength is normally distributed to decide whether Ahmad Fishing Line’s 10kg test line is not up to specifications. Perform the required hypothesis test at 5% significance level.

Answer 1

$$mean=\bar{x}=\dfrac{1}{12}(9.8+ 10.2+ 9.8+ 9.8 +9.6$$

$$+9.1+ 9.7 +9.7 +10.1+ 9.4 +10.1+ 9.7)$$

$$=9.75$$ $$s^2=\dfrac{1}{12-1}((9.8-9.75)^2+ (10.2-9.75)^2$$

$$+(9.8-9.75)^2+(9.8-9.75)^2+(9.6-9.75)^2$$

$$+(9.1-9.75)^2+(9.7-9.75)^2+(9.7-9.75)^2$$

$$+(10.1-9.75)^2+(9.4-9.75)^2+(10.1-9.75)^2$$

$$+(9.7-9.75)^2)=\dfrac{1.01}{11}\approx0.093636$$

$$s=\sqrt{s^2}=\sqrt{\dfrac{1.01}{11}}\approx0.306$$

The following null and alternative hypotheses need to be tested:

$H_0:\mu\geq10$

$H_1:\mu<10$

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 $\alpha = 0.01,$ $df=n-1=11$degrees of freedom, and the critical value for a left-tailed test is $t_c = - 1.795885.$

The rejection region for this left-tailed test is $R = \{t:t < -1.795885\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{9.75-10}{0.306/\sqrt{12}}\approx-2.8301$$

Since it is observed that $t = -2.8301< -1.795885= t_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for left-tailed $df=11$degrees of freedom, $t = -2.8301$ is $p=0.008185,$ and since $p=0.008185<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the mean tensile strength is less than $10,$ at the $\alpha = 0.05$ significance level.