The quality-control manager at a light emitting diode (LED) factory needs to determine whether the mean life of a large shipment of LEDs is equal to 50,000 hours.  The population standard deviation is 1,000 hours.  A random sample of 64 LEDs indicates a sample mean of 49,875 hours

(a) At a 0.05 level of significance, is there evidence that the mean life is different from 50,000 hours? We accept the Null.

This problem is a hypothesis test for the population mean. We need to test whether the mean life of LEDs is different from 50,000 hours using the provided sample data. Since the population standard deviation is known and the sample size is large (64), we will use the **Z-test** for the hypothesis.

### Step-by-Step Solution:

#### (a) Formulate the hypotheses:
We are testing if the mean life of the LEDs differs from 50,000 hours, so we have a two-tailed test.

- **Null Hypothesis (\(H_0\))**: The mean life of the LEDs is 50,000 hours.
  \[
  H_0: \mu = 50,000
  \]
  
- **Alternative Hypothesis (\(H_1\))**: The mean life of the LEDs is different from 50,000 hours.
  \[
  H_1: \mu \neq 50,000
  \]

#### (b) Significance level:
The significance level is given as \(\alpha = 0.05\), which implies a 5% level of significance for the test.

#### (c) Test statistic formula:
Since we know the population standard deviation (\(\sigma = 1,000\)) and the sample size is large (\(n = 64\)), we can use the **Z-test** statistic:

\[
Z = \frac{\bar{X} - \mu_0}{\frac{\sigma}{\sqrt{n}}}
\]
Where:
- \(\bar{X} = 49,875\) (sample mean),
- \(\mu_0 = 50,000\) (hypothesized population mean),
- \(\sigma = 1,000\) (population standard deviation),
- \(n = 64\) (sample size).

Let's calculate the test statistic.

\[
Z = \frac{49,875 - 50,000}{\frac{1,000}{\sqrt{64}}}
\]
\[
Z = \frac{-125}{\frac{1,000}{8}} = \frac{-125}{125} = -1
\]

#### (d) Critical value:
Since this is a two-tailed test at the 0.05 level of significance, we need to find the critical Z-values for \(\alpha/2 = 0.025\) in each tail. From the standard normal distribution table, the critical Z-values are:

\[
Z_{\text{critical}} = \pm 1.96
\]

#### (e) Decision rule:
- If the test statistic \(Z\) falls outside the range \([-1.96, +1.96]\), we reject the null hypothesis \(H_0\).
- If the test statistic \(Z\) falls within the range \([-1.96, +1.96]\), we fail to reject the null hypothesis \(H_0\).

#### (f) Conclusion:
The calculated Z-value is \(-1\), which lies within the critical region of \([-1.96, +1.96]\). Therefore, we **fail to reject** the null hypothesis at the 0.05 level of significance.

#### Interpretation:
There is not enough evidence at the 0.05 significance level to conclude that the mean life of the LEDs is different from 50,000 hours. Thus, we accept the null hypothesis.

---

Do you need more details, or have any questions?

### Follow-up questions:
1. What are Type I and Type II errors in this context?
2. How would the conclusion change if the significance level was 0.01?
3. What would happen if the population standard deviation were unknown?
4. How does the sample size impact the outcome of the test?
5. What is the difference between a Z-test and a t-test?

**Tip:** In hypothesis testing, the choice of significance level affects the likelihood of making a Type I error (rejecting a true null hypothesis).

The quality-control manager at a light emitting diode (LED) factory needs to determine whether the mean life of a large shipment of LEDs is equal to 50,000 hours. The population standard deviation is 1,000 hours. A random sample of 64 LEDs indicates a sample mean of 49,875 hours. At a 0.05 level of significance, is there evidence that the mean life is different from 50,000 hours? We accept the Null.

Determine whether the mean life of LEDs is different from 50,000 hours using a Z-test at a 0.05 level of significance. Learn the steps for hypothesis testing, including formulation of null and alternative hypotheses, calculation of test statistics, and interpretation of results. Suitable for students in Grades 11-12 or undergraduate statistics courses.

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