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------------------------------------------------ Situation B: Mike's fitness tracker automatically collects data on [tex]x[/tex], the number of miles he walks each day. Using a random sample of 16 days, he finds [tex]\sum x = 100.8[/tex] and [tex]\sum x^2 = 640.86[/tex].

Question B1: Find the mean of his sample.

- A. 6.30 miles
- B. 5.10 miles
- C. 7.78 miles
- D. 5.67 miles
- E. 7.00 miles

Question B2: Find the standard deviation of his sample.

- A. 0.409 miles
- B. 0.505 miles
- C. 0.623 miles
- D. 0.561 miles
- E. 0.454 miles

Question B3: Mike's fitness tracker also records the number of calories he burns each day. Using the same sample of 16 days, the sample mean for calories burned was 2585, with a sample standard deviation of 147.5. Using this information, find a 95% confidence interval for the mean number of calories Mike burns each day. This confidence interval extends from:

- A. 2498.6 to 2671.4 calories
- B. 2522.0 to 2648.0 calories
- C. 2514.2 to 2655.8 calories
- D. 2506.4 to 2663.6 calories
- E. 2490.8 to 2679.2 calories

Question B4: Fill in the blank: We should conclude (with 95% confidence) that ___ lies within the confidence interval.

- A. 95% of the population of Mike's daily calories burned
- B. 95% of the sample of Mike's daily calories burned
- C. the mean of the population of Mike's daily calories burned
- D. the mean of this sample of Mike's daily calories burned

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.

To learn more about interval click here:

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