Question

A supplier of USB flash claims that more than 11% of the USBs are defective. In a random sample of 167 USBs, it is found that 3 are defective, but the supplier claims that this is only a sample.

Answer 1

The claim that more than 11% of the USBs are defective is not supported by the sample data. The evidence suggests that the proportion of defective USBs is lower than the supplier's claim.

Step-by-step explanation:

The supplier's assertion that more than 11% of USBs are defective implies a population proportion greater than 0.11. The sample data, however, shows that in a random sample of 167 USBs, only 3 are defective. To assess the validity of the supplier's claim, we can perform a hypothesis test for a population proportion.

Firstly, we set up the null hypothesis (H0) that the proportion of defective USBs is 11% or less (p ≤ 0.11) and the alternative hypothesis (H1) that the proportion is greater than 11% (p > 0.11). Using statistical techniques, we can calculate a test statistic and compare it to a critical value or p-value.

In this case, the low number of defective USBs in the sample suggests that the evidence does not support the supplier's claim. The p-value associated with the test is likely to be greater than the significance level, leading us to fail to reject the null hypothesis. Consequently, we do not have sufficient evidence to conclude that the proportion of defective USBs is more than 11%.

In conclusion, based on the sample data, it is reasonable to question the supplier's claim that more than 11% of USBs are defective, as the evidence from the sample does not substantiate this assertion.

Answer 2

The evidence from the sample does not provide sufficient support to reject the supplier's claim. The sample proportion of defective USBs is below the claimed defect rate of 11%.

Step-by-step explanation:

The supplier claims that more than 11% of the USBs are defective. To test this claim, we can use a hypothesis test for a population proportion. Let (p) be the proportion of defective USBs. The null hypothesis
`\( H_0 \)` is that
`\( p \)` is less than or equal to 0.11, and the alternative hypothesis
`\( H_1 \)` is that
`\( p \)` is greater than 0.11.

In this case, the sample proportion
`\( \hat{p} \)`, which is the number of defective USBs divided by the sample size, is
`\( (3)/(167) \)`, approximately 0.0179. The standard error of
`\( \hat{p} \)` is calculated as
`\( \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \),` where ( n) is the sample size. Using the sample data, the standard error is found to be approximately 0.0166.

We can then use a Z-test to determine the p-value. The Z-statistic is calculated as
`\( \frac{\hat{p} - p_0}{\text{SE}(\hat{p})} \)` , where
`\( p_0 \)` is the hypothesized population proportion under the null hypothesis. In this case, the Z-statistic is approximately 0.761. Using the Z-statistic, we find the p-value, which is the probability of observing a sample proportion as extreme as
`\( \hat{p} \)` if the null hypothesis is true.

Comparing the p-value to a significance level (e.g., 0.05), if the p-value is less than the significance level, we reject the null hypothesis. In this case, the p-value is greater than 0.05, so we fail to reject the null hypothesis. Therefore, there is not enough evidence to support the supplier's claim that more than 11% of the USBs are defective based on the given sample.