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Anna11 [10]
3 years ago
15

A men’s health magazine would like to conduct a survey using a simple random sample of 900 men who practice yoga. One of the goa

ls of the survey is to create a confidence interval to estimate the proportion of all men practicing yoga who report low stress levels. Since the intended survey method requires face-to-face interaction, some of the magazine executives are concerned about the cost of the survey and suggest sampling a smaller group of 100 men instead of 900 men. Assuming the sample proportion doesn’t change, which of the following best describes the anticipated effect on the width of the confidence interval if the magazine were to survey a random sample of 100 men who practice yoga, rather than 900 men? A) The width of the interval based on 100 men would be about nine times the width of the interval based on 900 men. B) The width of the interval based on 100 men would be about three times the width of the interval based on 900 men. C) The width of the interval based on 100 men would be about the same width as the interval based on 900 men. D) The width of the interval based on 100 men would be about one-third the width of the interval based on 900 men. E) The width of the interval based on 100 men would be about one-ninth the width of the interval based on 900 men.
Advanced Placement (AP)
1 answer:
saw5 [17]3 years ago
4 0

Answer: B) The width of the interval based on 100 men would be about three times the width of the interval based on 900 men.

Explanation:

Based on the information given, if there's no change in the sample proportion, the option that best describes the anticipated effect on the width of the confidence interval assuming the magazine surveyed a random sample of 100 men who practice yoga, rather than 900 men will be option B "The width of the interval based on 100 men would be about three times the width of the interval based on 900 men.

Therefore, the correct option is B.

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Therefore the patient's average daily intake of sodium for those 4 days = 2542.5 mg

What is the patient’s average daily intake of sodium for those 4 days?

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given that mean average = 2300; we can also determine the fifth day sodium intake as follows:

mean average = \frac{2150+3005+2700+2315+y}{5}\\2300 = \frac{2150+3005+2700+2315+y}{5}

2300*5 =2150+3005+2700+2315+y

11500 = 10170 +y

11500-10170 =y

y = 1330 mg

Hence, his sodium intake for the fifth day will be 1330 in order to drop him below an average of 2300 mg

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