Question

A manufacturer of automobile seats has a production line that produces an average of 100 seats per day. Because of new Government regulations, new safety devices have been installed, which the manufacturer believes will reduce average daily output. A random sample of 15 days output after the installation of the safety devices is shown below:

93, 103, 95, 101, 91, 105, 96, 94, 101, 88, 98, 94, 101, 92, 95

Assuming that the daily output is normally distributed, is there sufficient evidence at the 5% level of significance, to conclude that average daily output has decreased following the installation of the safety devices?

Answer 1

Mean

$$\bar{x}=\dfrac{1}{15}(93+103+95+101+91+105$$

 

$$+96+94+101+88+98+94$$

$$+101+92+95)=\dfrac{1447}{15}\approx96.4667$$

Variance

$$s^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n-1}= 23.552381$$

$$s=\sqrt{s^2}\approx4.85308$$

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge100$

$H_a:\mu<100$

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 $\alpha = 0.05,$ $df=n-1=14$ degrees of freedom, and the critical value for a left-tailed test is $t_c = -1.76131.$

The rejection region for this left-tailed test is$R = \{t: t < -1.76131\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{96.4667-100}{4.85308/\sqrt{15}}\approx-2.81974$$

Since it is observed that $t = -2.81974 <-1.76131= t_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for left-tailed, $df=14$ degrees of freedom, $t=-2.81974$ is $p = 0.00682,$ and since $p=0.00682<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is less than 100, at the $\alpha = 0.05$ significance level.