Suppose a different police department says their SRS resulted in a 95% confidence interval of 10.1% to 15.1%. What is their point estimator and margin of error?
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Suppose a different police department says their SRS resulted in a 95% confidence interval of 10.1% to 15.1%. What is their point estimator and margin of error?
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- A small business wanted to estimate the mean income from their customers. They took a sample of 256 customers which yielded a sample mean of $71,250 with a standard deviation of $5,800 . What is the 90% confidence interval estimate for the population mean income of their customers.An internet retailer would like to investigate the relationship between the amount of time in minutes a purchaser spends on its Web site and the amount of money he or she spends on an order. The table to the right shows the data from arándome sample of 12 customers. Construct a 95% confidence interval for the regression slope. Construct a 95% confidence interval for the slope. LCL= UCL= I need help with this question step by step ! I have tried over and over.We have developed a new drug used to reduce fever and want to test its efficacy. Ten ER patients who are running a fever agree to take the new drug instead of the traditional treatment. Their temperatures are taken before the drug is administered and 30 minutes after taking the drug. The results are recorded in the Excel file below. Use the data to construct and interpret a 99% confidence interval for the mean change in temperatures of patients after taking this drug using Excel. Be sure to include any explanation necessary along with a sentence that explains what the interval means. First Temp Second Temp 100.1 98.9 101.3 99.1 102.1 99.2 102.7 99 101.9 98.7 100.8 98.6 103.1 99.4 102.5 99.2 103.5 100.1 101.7 99.2
- Take an SRS where n equals 50 and calculate at 95% confidence interval. If we took a different SRS of the same size from the same population, and calculate a confidence interval with the same level of confidence, the margin of error would stay the same. Is this statement True or False and explain why it is true or false.According to the website www.collegedrinkingprevention.gov, "About 25 percent of college students report academic consequences of their drinking including missing class, falling behind, doing poorly on exams or papers, and receiving lower grades overall." A statistics student is curious about drinking habits of students at his college. He wants to estimate the mean number of alcoholic drinks consumed each week by students at his college. He plans to use a 95% confidence interval. He surveys a random sample of 66 students. The sample mean is 3-75 alcoholic drinks per week. The sample standard deviation is 3.68 drinks. Construct the 95% confidence interval to estimate the average number of alcoholic drinks consumed each week by students at this college. Your answer should be rounded to 2 decimal places.Ms. Rains believes more seniors from PLD plan to attend college than seniors at Lafayette. In a random sample of 35 PLD seniors, 21 responded they plan to attend college. In a random sample of 32 Lafayette seniors only 17 responded they plan to attend college. Using a 95% confidence interval, predict the true difference in the proportion of seniors at PLD and Lafayette who plan to attend college and comment on Ms. Rains' belief.
- The Woodcock-Johnson Cognitive Ability test indicates that the average score is 95. The same group of 12th grade students (n = 72) in Columbus are tested and the sample mean is 110.1 and the sample standard deviation is 15.2. Do the scores provide good evidence that the mean IQ of this population is greater than 95? State the null (H0) and alternative (Ha) hypotheses. What are the confidence intervals for this data? What is the value of the test statistic? What is the p-value for this data? What is your decision about the null hypothesis? summerize the conclusion in the contextSuppose the police are able to take a SRS of cars and find that 12.3% of the cars in their sample have a safety violation with a margin of error of 2% for a 95% confidence interval. What is the confidence interval?In a survey of 508 adults, 317 indicated that emails can be easily misinterpreted. If we wish to construct a 95% confidence interval to estimate the true proportion of adults who believe that emails can be easily misinterpreted we would substitute a. P hat is equal to b. Z is equal to