Make a 99 confidence interval around a mean of 229 with a


Please help answer the following question

1) Some research suggests that police officers are more likely to make an arrest in the presence of bystanders. If mentally disordered suspects attract more attention from bystanders, they may be more likely to be arrested than nonmentally disordered suspects. Assume that the average number of bystanders when police encounter a suspect is 3.58. In a sample of 20 police encounters with mentally disordered suspects, Engel and Silver (2001) found an average of 7.03 bystanders (standard deviation = 9.42). Perform a hypothesis test to determine if there are more bystanders when the suspect is mentally disordered. Use a 0.01 level of statistical significance.

What's your critical value?

2) Make a 99% confidence interval around a mean of 22.9 with a standard deviation of 1.9, with a sample size of 18. For this confidence interval, what would be the critical value you'd use?

3) You've conducted a hypothesis test and now you're at the last step: the conclusion. You've for a test statistic of t = -0.256 and a critical value of -3.268 (i.e., a one tailed test). Based on this information, what would you conclude?

4) Some research suggests that police officers are more likely to make an arrest in the presence of bystanders. If mentally disordered suspects attract more attention from bystanders, they may be more likely to be arrested than nonmentally disordered suspects. Assume that the average number of bystanders when police encounter a suspect is 3.58. In a sample of 20 police encounters with mentally disordered suspects, Engel and Silver (2001) found an average of 7.03 bystanders (standard deviation = 9.42). Perform a hypothesis test to determine if there are more bystanders when the suspect is mentally disordered. Use a 0.05 level of statistical significance.

What is your test statistic?

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Basic Statistics: Make a 99 confidence interval around a mean of 229 with a
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