Question

an electronics manufacturing process has historically had a mean completion time of 70 minutes. it is claimed that, due to improvements in the process, the mean completion time, μ, is now less than 70 minutes. a random sample of 9 completion times using the new process is taken. the sample has a mean completion time of 69 minutes, with a standard deviation of 11 minutes.
assume that completion times using the new process are approximately normally distributed. at the 0.05 level of significance, can it be concluded that the population mean completion time using the new process is less than 70 minutes?
perform a one - tailed test. then complete the parts below.
carry your intermediate computations to three or more decimal places. (if necessary, consult a list of formulas.)
(a) state the null hypothesis (h_0) and the alternative hypothesis (h_1).
(h_0:)
(h_1:)
(b) determine the type of test statistic to use.
select
(c) find the value of the test statistic. (round to three or more decimal places.)

(d) find the p - value. (round to three or more decimal places.)

(e) can it be concluded that the mean completion time using the new process is less than 70 minutes?
yes no

Explanation:

Step1: State hypotheses

The null hypothesis $H_0$ is the status - quo, and the alternative hypothesis $H_1$ is what we are trying to find evidence for. Since we want to test if the mean is less than 70, $H_0:\mu = 70$ and $H_1:\mu<70$.

Step2: Determine test statistic type

The population standard - deviation $\sigma$ is unknown, and we have a sample standard - deviation $s$. Also, the sample size $n = 9$ and the population is approximately normally distributed. So, we use a t - test statistic. The formula for the t - test statistic is $t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$.

Step3: Calculate the test statistic

We are given $\bar{x}=69$, $\mu = 70$, $s = 11$, and $n = 9$.
\[

$$\begin{align*} t&=\frac{\bar{x}-\mu}{s/\sqrt{n}}\\ &=\frac{69 - 70}{11/\sqrt{9}}\\ &=\frac{- 1}{11/3}\\ &=\frac{-1}{\frac{11}{3}}\approx - 0.273 \end{align*}$$

\]

Step4: Calculate the p - value

The degrees of freedom is $df=n - 1=9 - 1 = 8$. For a one - tailed t - test with $t=-0.273$ and $df = 8$, using a t - distribution table or software, the p - value is $P(t_{8}<-0.273)\approx0.397$.

Step5: Make a decision

The significance level is $\alpha = 0.05$. Since the p - value ($0.397$) is greater than $\alpha(0.05)$, we fail to reject the null hypothesis.

Answer:

(a) $H_0:\mu = 70$, $H_1:\mu<70$
(b) t - test statistic
(c) $-0.273$
(d) $0.397$
(e) No