Question

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A data set includes 103 body temperatures of healthy adult humans having a mean of 98.5°F and a standard deviation of 0.61". Construct a 99% confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of 98.6°F as the mean body temperature? Click here to view at distribution table. Click here to view page 1 of the star dard normal distribution table. Click here to view page 2 of the standard normal distribution table What is the confidence interval estimate of the population mean? F<<DF (Round to three decimal places as needed.) What does this suggest about the use of 98.6°F as the mean body temperature? O A. This suggests that the mean body temperature is significantly lower than 98.65 O B. This suggests that the mean body temperature could very possibly be 98.69 Click to select your answer(s). Save for Later

Answer 1

i)

Let T denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here, 0.610 C= 98.500, s2 = 0.372100, s = 0.610, n = 103, standard error 0.060105 0.060 V103 The critical value is tcritical t 1-0.99 ,102 = 50.005,102 = 2.624891 2.625 At 99 percent confidence level, The margin of error is: S 0.610 ME= * t 1-0.99 ,102 2 0.610 V103 * +0.005,102 * 2.625 = 0.157990 0.158 V103 a 99 percent confidence interval of population mean u is S *t 1-0.99 == ✓n -,102 0.610 = 98.500 € *t0.005,102 103 = 98.500 + 0.610 V103 * 2.625 = 98.500 + 0.158 = (98.342010,98.657990) ~ (98.342,98.658)

ii) Since 98.6 lies within the confidence interval, so we can say that -

B) This suggests that the mean body temperature could very possibly be 98.6 F.