A sample of 900900 computer chips revealed that 25%25% of the chips do not fail in the first 10001000 hours of their use.  The company's promotional literature states that 28%28% of the chips do not fail in the first 10001000 hours of their use.  The quality control manager wants to test the claim that the actual percentage that do not fail is different from the stated percentage.  Is there enough evidence at the 0.020.02 level to support the manager's claim?

To solve this problem, we will perform a hypothesis test for proportions using the following steps:

### Step 1: Set up the hypotheses
We need to test whether the actual percentage of computer chips that do not fail in the first 1000 hours is different from the claimed percentage of 28%. The null and alternative hypotheses are:

\[
H_0: p = 0.28 \quad \text{(The true proportion is 28%)}
\]
\[
H_1: p \neq 0.28 \quad \text{(The true proportion is different from 28%)}
\]

This is a two-tailed test.

### Step 2: Define the significance level
The significance level is given as \( \alpha = 0.02 \).

### Step 3: Calculate the test statistic
The sample proportion is given by:

\[
\hat{p} = \frac{225225}{900900} = 0.25
\]

The test statistic for a proportion is calculated using the formula:

\[
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}}
\]

Where:
- \( \hat{p} = 0.25 \) is the sample proportion,
- \( p_0 = 0.28 \) is the claimed proportion,
- \( n = 900900 \) is the sample size.

### Step 4: Calculate the critical value and p-value
Since this is a two-tailed test, we will compare the calculated test statistic to the critical z-value for a significance level of 0.02.

Let's compute the test statistic first. I'll calculate the value.The calculated test statistic is approximately \( z = -63.42 \).

### Step 5: Determine the critical z-value
For a two-tailed test at a significance level of \( \alpha = 0.02 \), the critical z-values correspond to the upper and lower 1% tails of the standard normal distribution. The critical z-value for a two-tailed test at \( \alpha = 0.02 \) is approximately \( \pm 2.33 \).

### Step 6: Make a decision
Since the calculated \( z \)-value of \( -63.42 \) is far outside the critical range of \( -2.33 \) to \( +2.33 \), we reject the null hypothesis.

### Conclusion:
There is enough evidence at the 0.02 significance level to support the manager's claim that the actual percentage of chips that do not fail in the first 1000 hours is different from the stated percentage of 28%.

Would you like more details, or do you have any questions?

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Here are 5 related questions:
1. How do you interpret the p-value in hypothesis testing?
2. What are Type I and Type II errors in hypothesis testing?
3. How would a one-tailed test change the hypothesis setup in this example?
4. How does the sample size affect the test statistic in hypothesis tests for proportions?
5. How would the conclusion change if the significance level were 0.05 instead of 0.02?

**Tip:** A larger sample size generally provides more precise estimates, leading to a smaller margin of error in hypothesis testing.

A sample of 900900 computer chips revealed that 25% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature states that 28% of the chips do not fail in the first 1000 hours of their use. The quality control manager wants to test the claim that the actual percentage that do not fail is different from the stated percentage. Is there enough evidence at the 0.02 level to support the manager's claim?

This problem tests whether the actual proportion of computer chips that do not fail within the first 1000 hours differs from the company's claimed 28%. Using a hypothesis test for proportions at a 0.02 significance level, learn how to calculate the z-value and make a statistical decision. Suitable for college-level statistics students.

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