The table contains the percentage of woman receiving prenatal care for a sample of countries. Estimate the average percentage of woman receiving prenatal care using a 90% confidence interval. Round to two decimal places.
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- Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. Of 92 adults selected randomly from one town, 61 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance. A) 0.548 < p < 0.778 B) 0.536 < p < 0.790 C) 0.582 < p < 0.744 D) 0.566 < p < 0.760The owner of the AGT Gas Station wishes to determine the fraction of customers who pay using a credit card or debit card. A random sample of 100 customers was surveyed and finds that 80 paid using their credit card or debit card. If you are going to construct a 99% confidence interval for the fraction of customers who paid using their credit card or debit card, what will be the value of the margin of error? a. 0.0658 b. 0.103 c. 0.0784 d. 0.0299A random sample of 34 participants in a Zumba dance class had their heart rates measured before and after a moderate 10-minute workout. The following data correspond to the increase in each individual’s heart rate (in beats per minute): 59 70 57 42 57 59 41 54 44 36 59 61 52 42 41 32 60 54 52 53 51 47 62 62 44 69 50 37 50 54 48 52 61 45 What is the point estimate of the corresponding population mean? Make a 98% confidence interval for the average increase in a person’s heart rate after a moderate 10-minute Zumba workout. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Describe all possible alternatives. Which alternative is the best and why?