Question

A manufacturer of automobile seats has a production line that produces an average of
100 seats per day. Because of new government regulations, a new safety device has
been installed, which the manufacturer believes will reduce average daily output. A
random sample of 15 days output after the installation of the safety device is shown:
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 to
conclude that average daily output has decreased following the installation of the
safety device? (Use α = 0.05 )

Answer 1

Answer on Question #66220 - Math - Statistics and Probability

Question

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

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 to conclude that average daily output has decreased following the installation of the safety device?

(Use $\alpha = 0.05$)

Solution

The null hypothesis is $H_0: \mu \geq 100$.

The alternative hypothesis is $H_1: \mu < 100$.

Since the claim $\mu < 100$ does not contain equality, it is an alternative hypothesis about the population mean.

Since the distribution is normal, but the population standard deviation is unknown, we will use $t$-statistic to test the null hypothesis.

The sample size is $n = 15$. The sample mean equals

$$\bar{x} = \frac{\sum (\text{daily output})}{\text{days}} = 96.46667.$$

The standard deviation of the sample equals

$$s = \sqrt{\frac{\sum ((\text{daily output} - \bar{x})^2)}{(\text{days} - 1)}} = 4.85308.$$

The observed value of the statistic equals

$$\bar{t} = \frac{96.46667 - 100}{4.85308 / \sqrt{15}} = -2.81976.$$

The degree of freedoms equals

$$df = 15 - 1 = 14.$$

The significant level is $\alpha = 0.05$.

Method 1

The critical value for this left-tailed test is

$$- t_{14,0.05} = -1.761.$$

Since $\bar{t} < -t_{14,0.05}$, we conclude that there is sufficient evidence at the 0.05 level of significance to reject the claim that the daily output has not decreased.

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Method 2

The p-value for this test is

$$p = P(t < -2.81976) = 0.00682.$$

Since $p < \alpha$, we conclude that there is sufficient evidence at the 0.05 level of significance to reject the claim that the daily output has not decreased.

Answer:

Yes. There is a sufficient evidence to conclude that average daily output has decreased following the installation of the safety device.