The resting heart rates for 80 women aged 46–55 in a simple random sample are normally distributed, with a mean of 71 beats per minute and a standard deviation of 6 beats per minute. Assuming a 90% confidence level, what is the margin of error for the population mean? 0.66 1.10 1.31 1.73
2 answers:
Answer:
the answer is b. 1.10
Step-by-step explanation:
i just took this test, its right
Given: Confidence level = 90% mean = 71 beats per minute standard deviation = 6 beats per minute margin of error = z * δ / √n where : δ - population of the standard deviation, n is the sample size ; z is the appropriate z value. 90% confidence level = 1.645 in z-value margin of error = 1.645 * (6/√80) = 1.645 * (6/8.94) = 1.645 * 0.671 = 1.104The margin of error is 1.10
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