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Assignment Instructions Let X = body temperature in degrees Fahrenheit of a randomly selected person. Does X meet the criteria for a random variable? Explain.

Assignment Instructions

Let X = body temperature in degrees Fahrenheit of a randomly selected person. Does Xmeet the criteria for a random variable? Explain. If so, would Xbe a discrete or continuous random variable?

  1. If human body temperatures in the United States are normally distributed with a mean of 97.5 and standard deviation of 0.7, calculate P(X > 98.6). Would it be considered unusual for an individual to have a body temperature above 98.6 degrees? Show your work, including calculator input, function, and output.
  2. We want to construct a 95% confidence interval for the mean body temperature of college students. Consider the sample of body temperatures we collected at the beginning of the semester. Have the assumptions to construct a confidence interval been met? Explain.

Pretend our sample data values were from a simple random sample.

a. Construct a 95% confidence interval for the mean body temperature of college students using our sample of student data. Show all your work including calculator input, calculator function and output. b. Use a complete sentence to interpret this confidence interval. c. Recent studies suggests that the mean body temperature of humans is less than 98.6 F. Based on the confidence interval, is it reasonable that the mean body temperature of college students is less than 98.6? Explain.

  1. If we use a previous study that suggests = 0.7 Fahrenheit for the population of college students in the United States, what sample size is necessary to construct a 95% confidence interval for the body temperature of college students so that it will have a margin of error of 0.2 F?

Suppose our sample data values were from a simple random sample.

a. Using the temperature data set, test the claim that < 98.6 F. Be sure to include all steps for conducting a hypothesis test. Show all your work including calculator input, calculator function and output. Use =0.05 level of significance and the p-value method.

b. Use a complete sentence to interpret the results.

Data:

97.4, 95.5, 98.9, 98.2, 98, 98.7, 97.2, 97.7, 97.9, 97.4, 96.9, 96.4, 99.2, 98.5, 98.3, 98.4, 98.3, 98.6, 98.9, 97.8, 97.4, 99.3, 98.6, 98, 97.3, 98.2, 97.3, 97.6, 97.5, 97.1, 99, 98.6, 98.7, 98.8

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