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# There were no defectives in a sample of 1 light bulb does this sample provide sufficient evidence that in the warehouse with millions of light bulbs fewer than 10% are defective?

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## Answer to a math question There were no defectives in a sample of 1 light bulb does this sample provide sufficient evidence that in the warehouse with millions of light bulbs fewer than 10% are defective?

Clarabelle
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To test the claim that fewer than 10% of light bulbs in the warehouse are defective, we can use a hypothesis test for a population proportion. The steps are as follows: - Check the conditions: The sample is randomly selected, there are only two options $defective or not defective$, and the sample has at least 5 members in each category. In this case, the sample size is 1, which is too small to meet the conditions. Therefore, we cannot use a hypothesis test for a proportion with this sample. - Define the claims: The null hypothesis is that the population proportion of defective light bulbs is equal to 10%, and the alternative hypothesis is that it is less than 10%. In symbols, we have: H0: p = 0.10 Ha: p < 0.10 This is a left-tailed test, because the alternative hypothesis claims that the proportion is less than in the null hypothesis. - Decide the significance level: This is the probability of rejecting the null hypothesis when it is true. A common choice is 0.05, but it can vary depending on the context and the consequences of making a wrong decision. - Calculate the test statistic: This is a measure of how far the sample proportion is from the hypothesized proportion, relative to the standard error of the sampling distribution. The formula is: z = $p' - p0$ / sqrt$p0 * (1 - p0$ / n) where p' is the sample proportion, p0 is the hypothesized proportion, and n is the sample size. In this case, we have: z = $0 - 0.10$ / sqrt$0.10 * (1 - 0.10$ / 1) = -0.333 - Calculate the p-value: This is the probability of obtaining a test statistic as extreme or more extreme than the observed value, assuming the null hypothesis is true. For a left-tailed test, the p-value is the area to the left of the test statistic on the standard normal curve. We can use a calculator or a table to find the p-value. In this case, we have: p-value = P$z < -0.333$ = 0.369. The p-value equals 0.369. - Make a decision: We compare the p-value to the significance level, and reject the null hypothesis if the p-value is smaller. In this case, we have: p-value < significance level 0.369 > 0.05 To make a decision based on the p-value, we need to compare it to a significance level, which is the probability of rejecting the null hypothesis when it is true. A common choice for the significance level is 0.05, but it can vary depending on the context and the consequences of making a wrong decision. If the p-value is less than or equal to the significance level, then we reject the null hypothesis and conclude that there is sufficient evidence to support the alternative hypothesis. If the p-value is greater than the significance level, then we fail to reject the null hypothesis and conclude that there is not enough evidence to support the alternative hypothesis. In this case, the p-value is 0.369, which is greater than 0.05, so we would fail to reject the null hypothesis and say that the data does not provide enough evidence to support the claim that fewer than 10% of light bulbs in the warehouse are defective. However, this conclusion is based on a very small sample size, which may not be representative of the population. A larger sample would provide more reliable results and reduce the sampling error.
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