Final answer:
Mike would use a chi-square test of independence at a 5% level of significance to determine if there's enough evidence of a difference in breakout frequencies between prison wings. If a 1% significance level is used, the outcome might differ. Also, Mike's sample may suffer from sampling bias as it only represents the last ten years and might not reflect long-term trends.
Step-by-step explanation:
To assess whether there is enough evidence to conclude that, over the 100-year period, breakouts were more likely to occur from certain wings of Didsbury Prison than from others, Mike would need to perform a chi-square test bindependence. Using the data for the past ten years with a total of 30 breakouts and the distribution across the wings, Mike can conduct this test at a 5% level of significance. The steps would include:
- Setting up the null hypothesis (H0) that there is no difference in breakout frequency across the wings.
- Setting up the alternative hypothesis (H1) that there is a difference in breakout frequency across the wings.
- Calculating the expected frequencies for each wing based on uniform distribution.
- Computing the chi-square statistic using observed and expected frequencies.
- Comparing the chi-square statistic to the critical value from the chi-square distribution, or obtaining the p-value.
- Determining if the evidence is strong enough to reject the null hypothesis.
If a 1% level of significance were used, the critical value would be higher, which could potentially change the conclusion if the chi-square statistic was borderline significant at the 5% level.
As for problems with Mike's choice of a sample, it may not be representative of the entire 100-year period due to changes in prison security, administration, or the physical state of the wings over time. This could lead to a sampling bias. Furthermore, only considering the last ten years might not accurately reflect the overall trends and variances that could have taken place over the entire century.