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, , is now less than 70 minutes. A random sample of 16 completion times using the new process is taken. The sample has a mean completion time of 69 minutes, with a standard deviation of 12 minutes.
Assume that completion times using the new process are approximately normally distributed. At the 0.10 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. (a)State the null hypothesis H0 and the alternative hypothesis H1. (b)	Determine the type of test statistic to use. (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

To perform a hypothesis test for the given scenario, we'll go through the steps systematically. Here's how we approach the problem:

### (a) State the Null Hypothesis \( H_0 \) and the Alternative Hypothesis \( H_1 \)

- **Null Hypothesis \( H_0 \)**: The mean completion time using the new process is equal to 70 minutes.
  \[
  H_0: \mu = 70 \text{ minutes}
  \]

- **Alternative Hypothesis \( H_1 \)**: The mean completion time using the new process is less than 70 minutes.
  \[
  H_1: \mu < 70 \text{ minutes}
  \]

### (b) Determine the Type of Test Statistic to Use

Since the population standard deviation is not known and the sample size is small (\(n = 16\)), we use the **t-test** statistic.

### (c) Find the Value of the Test Statistic

The test statistic for a t-test is calculated using the formula:
\[
t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}
\]
where:
- \(\bar{x}\) is the sample mean,
- \(\mu_0\) is the population mean under the null hypothesis,
- \(s\) is the sample standard deviation,
- \(n\) is the sample size.

Substituting the values:
\[
t = \frac{69 - 70}{\frac{12}{\sqrt{16}}} = \frac{-1}{\frac{12}{4}} = \frac{-1}{3} = -0.333
\]

So, the value of the test statistic is:
\[
t = -0.333
\]

### (d) Find the p-Value

To find the p-value, we need to determine the probability that the test statistic is less than the calculated value. Since this is a one-tailed test, we look at the cumulative probability for \( t = -0.333 \) with \( df = n - 1 = 15 \) degrees of freedom.

Using a t-distribution table or calculator, we find:
\[
\text{p-value} \approx 0.370
\]

### (e) Conclusion: Can It Be Concluded that the Mean Completion Time Using the New Process Is Less than 70 Minutes?

To determine whether to reject the null hypothesis, compare the p-value to the significance level \(\alpha = 0.10\).

- Since \( \text{p-value} \approx 0.370 \) is **greater** than \( \alpha = 0.10 \), we **fail to reject** the null hypothesis.

**Conclusion: NO, it cannot be concluded that the mean completion time using the new process is less than 70 minutes.**

---

Do you want more details or have any other questions?  
Here are some related questions you might find useful:

1. What if the sample size was larger? How would that impact the test?
2. How do you interpret a p-value in the context of hypothesis testing?
3. What is the relationship between the p-value and the significance level?
4. How does the choice of significance level affect the outcome of a hypothesis test?
5. What are the assumptions behind using a t-test for hypothesis testing?
6. How would the test differ if the population standard deviation was known?
7. Can you explain the concept of a one-tailed versus a two-tailed test?
8. What is the importance of the degrees of freedom in a t-test?

**Tip:** When conducting hypothesis tests, it's important to ensure that the assumptions underlying the test are met (e.g., normality of the data, appropriate sample size). This ensures the validity of the test results.

Explore a hypothesis test for mean completion time in electronics manufacturing. Perform a one-tailed t-test with detailed steps and calculations. Given a sample mean of 69 minutes and a standard deviation of 12 minutes, determine if the mean completion time using the new process is less than 70 minutes.

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