In a manufacturing plant, if the manager believes that more than [tex]$5\%$[/tex] of cans are dented, he will have the machines shut down for repair at great cost. Otherwise, he will let them continue to run.

To make the decision, the manager will inspect a random sample of 100 cans, then perform a test at the [tex]$\alpha=0.05$[/tex] significance level of [tex]$H_0: p=0.05$[/tex] versus [tex]$H_a: p\ \textgreater \ 0.05$[/tex] where [tex]$p$[/tex] is the true proportion of all cans that are dented.

- Consequence of a Type I error:
- The manager will have the machines shut down for repair at great cost, unnecessarily.
- Consequence of a Type II error:
- The manager will let the machines continue to run, which will continue to damage more than [tex]$5\%$[/tex] of the cans. Profit will be lost, and the machines will still need to be repaired.



Answer :

### Solution:

#### Step 1: Define the Hypotheses
We start by defining the null and alternative hypotheses:
- Null Hypothesis [tex]\( H_0 \)[/tex]: [tex]\( p = 0.05 \)[/tex] (The true proportion of dented cans is 5%)
- Alternative Hypothesis [tex]\( H_a \)[/tex]: [tex]\( p > 0.05 \)[/tex] (The true proportion of dented cans is greater than 5%)

#### Step 2: Sample Size and Significance Level
We have a sample size [tex]\( n = 100 \)[/tex] and a significance level of [tex]\( \alpha = 0.05 \)[/tex].

#### Step 3: Sample Proportion
The sample proportion of dented cans is given as 5%, or [tex]\( \hat{p} = 0.05 \)[/tex].

#### Step 4: Population Proportion Under the Null Hypothesis
Under the null hypothesis, the population proportion [tex]\( p_0 = 0.05 \)[/tex].

#### Step 5: Standard Deviation of the Sampling Distribution
The standard deviation for the sampling distribution of the sample proportion is calculated using the formula:
[tex]\[ \sigma_{\hat{p}} = \sqrt{\frac{p_0 (1 - p_0)}{n}} \][/tex]
Given the values:
[tex]\[ \sigma_{\hat{p}} = \sqrt{\frac{0.05 \times 0.95}{100}} \approx 0.0218 \][/tex]

#### Step 6: Calculate the Z-Score
The z-score for the sample proportion is calculated as:
[tex]\[ z = \frac{\hat{p} - p_0}{\sigma_{\hat{p}}} \][/tex]
Substituting the given values:
[tex]\[ z = \frac{0.05 - 0.05}{0.0218} = 0 \][/tex]

#### Step 7: Find the Critical Value
For a significance level of [tex]\( \alpha = 0.05 \)[/tex] and a right-tailed test, we need to find the critical value. The critical value for [tex]\( \alpha = 0.05 \)[/tex] is:
[tex]\[ z_{0.05} \approx 1.645 \][/tex]

#### Step 8: Compare the Z-Score to the Critical Value
We compare the calculated z-score to the critical value:
[tex]\[ z = 0 \quad \text{and} \quad z_{0.05} = 1.645 \][/tex]

Since [tex]\( 0 \leq 1.645 \)[/tex], we fail to reject the null hypothesis.

#### Conclusion
Since we fail to reject the null hypothesis, we conclude that there is not enough evidence to suggest that the proportion of dented cans is greater than 5%. Therefore, the manager should continue to run the machines.

### Implications of Errors
1. Type I Error (Rejecting [tex]\( H_0 \)[/tex] when it is true):
- Consequence: The manager will shut down the machines for repair at great cost, unnecessarily.
2. Type II Error (Failing to reject [tex]\( H_0 \)[/tex] when it is false):
- Consequence: The manager will let the machines continue to run, which will continue to damage more than 5% of the cans. Profit will be lost, and the machines will still need to be repaired.

By carefully setting up and interpreting this hypothesis test, the manager can make a more informed decision regarding whether to shut down the machines for repairs or not.