Answer :
The 95% confidence interval for the mean number of calories Mike burns each day is approximately from 2506.392 to 2663.608 calories.
B1: The mean of Mike's sample can be calculated by dividing the sum of the values (∑x) by the sample size (n). In this case, ∑x = 100.8 and the sample size is 16. Thus, the mean of his sample is 100.8/16 = 6.30 miles.
B2: The standard deviation of the sample can be calculated using the formula:
Standard Deviation (s) = √[(∑x^2 - (∑x)^2/n) / (n - 1)]
Substituting the given values, we have:
s = √[(640.86 - (100.8)^2/16) / (16 - 1)]
≈ √[(640.86 - 640.8) / 15]
≈ √[0.04 / 15]
≈ √0.00267
≈ 0.0517 miles
Therefore, the standard deviation of his sample is approximately 0.0517 miles.
B3: To find a 95% confidence interval for the mean number of calories Mike burns each day, we can use the formula:
Confidence Interval = sample mean ± (critical value) * (sample standard deviation / √n)
The sample mean is 2585, the sample standard deviation is 147.5, and the sample size is 16.
The critical value for a 95% confidence interval with a sample size of 16 can be obtained from the t-distribution table or calculator. In this case, it is approximately 2.131.
Plugging in the values, we get:
Confidence Interval = 2585 ± (2.131) * (147.5 / √16)
= 2585 ± 2.131 * 36.875
= 2585 ± 78.608
≈ 2506.392 to 2663.608 calories
Therefore, the 95% confidence interval for the mean number of calories Mike burns each day is approximately from 2506.392 to 2663.608 calories.
B4: We should conclude (with 95% confidence) that the mean of the population of Mike's daily calories burned lies within the confidence interval.
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