A manufacturer of flashlight batteries claims that the mean life of his product will exceed 30 hours. A company is willing to buy a very large quantity of batteries if the claim is true. A random sample of 36 batteries is tested and it is found that the sample mean is 34 hours, with a standard deviation of 5 hours. Would the company buy?

This problem can be analyzed using hypothesis testing to determine whether the company's claim that the mean life of the batteries exceeds 30 hours is statistically significant.

We will conduct a **one-sample t-test** for the mean, as the population standard deviation is unknown, and we are provided with a sample standard deviation.

### Steps for Hypothesis Testing:

1. **Null Hypothesis (H₀):**  
   \( H_0: \mu \leq 30 \) (The mean life of the batteries is less than or equal to 30 hours)

2. **Alternative Hypothesis (H₁):**  
   \( H_1: \mu > 30 \) (The mean life of the batteries is greater than 30 hours)

3. **Significance Level (α):**  
   A typical significance level is \( \alpha = 0.05 \), unless otherwise stated.

4. **Test Statistic (t-score):**  
   Since the sample size is 36 and we know the sample standard deviation, we will use the t-test formula for the test statistic:
   \[
   t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}
   \]
   Where:
   - \( \bar{x} = 34 \) (sample mean),
   - \( \mu_0 = 30 \) (hypothesized population mean),
   - \( s = 5 \) (sample standard deviation),
   - \( n = 36 \) (sample size).

5. **Decision Rule:**  
   Compare the computed t-value with the critical t-value from the t-distribution table for \( n-1 = 35 \) degrees of freedom at a significance level of 0.05 for a one-tailed test.

### Let's calculate the t-score.

\[
t = \frac{34 - 30}{\frac{5}{\sqrt{36}}} = \frac{4}{\frac{5}{6}} = \frac{4}{0.8333} \approx 4.8
\]

### Interpretation:

- The computed **t-value = 4.8**.
- For a one-tailed test at \( \alpha = 0.05 \) and \( df = 35 \), the critical t-value from the t-distribution is approximately **1.689**.

Since the computed t-value (4.8) is much greater than the critical t-value (1.689), we **reject the null hypothesis**.

### Conclusion:
There is sufficient evidence to support the claim that the mean life of the batteries exceeds 30 hours. Therefore, the company would likely buy the batteries based on the test results.

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Would you like more details on any part of this solution or clarification on how hypothesis testing works?

### Related questions:
1. What is the difference between a one-tailed and two-tailed test?
2. How do you calculate the p-value for this hypothesis test?
3. What assumptions are required for the t-test to be valid?
4. How would the decision change if the sample size were smaller?
5. What is the impact of changing the significance level on the result?

**Tip:** Always ensure that the assumptions of normality and random sampling are met when using a t-test for small sample sizes.

This page covers hypothesis testing using a one-sample t-test to determine if the mean battery life exceeds 30 hours. Given a sample mean of 34 hours and a standard deviation of 5 hours from a sample of 36 batteries, the test concludes the company would likely buy the batteries. Suitable for advanced high school or college students.

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