smith is a weld inspector at a shipyard. he knows from keeping track of good and substandard welds that for the afternoon shift 5% of all welds done will be substandard. If Smith checks 300 of the 7500 welds completed that shift, would it be unusual for Smith to find 30 or more substandard welds? Justify your answer
To determine whether it would be unusual for Smith to find 30 or more substandard welds out of the 300 welds he checked, we can use statistical analysis. Specifically, we can utilize probability theory and hypothesis testing.
Step 1: Set Up the Hypotheses
We need to establish two hypotheses: the null hypothesis (H0) and the alternative hypothesis (H1).
- Null Hypothesis (H0): The number of substandard welds found by Smith is within the expected range.
- Alternative Hypothesis (H1): The number of substandard welds found by Smith is not within the expected range, implying it is unusual.
Step 2: Define the Significance Level
The significance level, denoted as α (alpha), determines how confident we need to be to reject the null hypothesis. Common significance levels are 0.05 and 0.01. Let's assume α = 0.05 for this scenario.
Step 3: Calculate the Expected Number of Substandard Welds
According to the information provided, Smith knows that 5% of the total welds done will be substandard. Therefore, the expected number of substandard welds out of 300 can be calculated as follows:
Expected number of substandard welds = 0.05 * 300 = 15
Step 4: Perform the Test
To test whether finding 30 or more substandard welds is unusual, we can use a binomial distribution. We calculate the probability of finding 30 or more substandard welds under the assumption of the null hypothesis. If this probability is less than or equal to the significance level, we reject the null hypothesis.
Let's perform the calculation:
P(X ≥ 30) = P(X = 30) + P(X = 31) + ... + P(X = 300)
To calculate each individual probability, we need to use the binomial probability formula. However, doing this manually for all values would be time-consuming. An alternative way is to use statistical software or calculators with built-in binomial probability functions.
Step 5: Make a Decision
If the probability calculated in Step 4 is less than or equal to the significance level (α = 0.05), we reject the null hypothesis. This indicates that finding 30 or more substandard welds would be considered unusual.
Note: Since performing the actual calculations can be complex, it is recommended to use statistical software or online calculators capable of calculating such probabilities efficiently.
By following this process, Smith can determine whether finding 30 or more substandard welds would be unusual given their knowledge of the expected percentage of substandard welds.
To determine if it would be unusual for Smith to find 30 or more substandard welds, we can use statistical analysis.
First, let's calculate the expected number of substandard welds in the 300 welds that Smith checked. We know that 5% of all welds are substandard, so the expected number of substandard welds would be:
Expected number of substandard welds = (5/100) * 300 = 15
Next, we need to calculate the standard deviation of the number of substandard welds. The standard deviation can be calculated using the formula:
Standard deviation = sqrt(n * p * (1-p))
Where:
- n is the sample size (300 in this case)
- p is the probability of success (proportion of substandard welds, which is 5/100 = 0.05 in this case)
Standard deviation = sqrt(300 * 0.05 * (1-0.05)) = sqrt(300 * 0.05 * 0.95) = sqrt(7.125) ≈ 2.67
Now, let's determine the z-score of finding 30 substandard welds:
Z-score = (x - μ) / σ
Where:
- x is the observed number of substandard welds (30 in this case)
- μ is the expected number of substandard welds (15)
- σ is the standard deviation (2.67)
Z-score = (30 - 15) / 2.67 = 15 / 2.67 ≈ 5.62
Finally, we can determine the probability of finding 30 or more substandard welds using the z-score. We will use a z-table or a statistical software to find this probability. Assuming a normal distribution and rounding to four decimal places, a z-score of 5.62 corresponds to a probability of essentially 1 (or 100%).
Therefore, it would be unusual for Smith to find 30 or more substandard welds in the 300 welds that he checked, with a probability close to 0.
Yes.
5% of 300 = 15