Question

Over the past several months, an adult patient has been treated for tetany (severe muscle spasms). This condition is associated with an average total calcium level below 6 mg/dl. Recently, the patient's total calcium tests gave the following readings (in mg/dl). Assume that the population of x values has an approximately normal distribution.

9.3

9.0

10.7

9.3

9.4

9.8

10.0

9.9

11.2

12.1

(a) Use a calculator with mean and sample standard deviation keys to find the sample mean reading x and the sample standard deviation s. (Round your answers to two decimal places.)

x =	mg/dl
s =	mg/dl

(b) Find a 99.9% confidence interval for the population mean of total calcium in this patient's blood. (Round your answer to two decimal places.)

lower limit	mg/dl
upper limit	mg/dl

(c) Based on your results in part (b), do you think this patient still has a calcium deficiency? Explain.

Yes. This confidence interval suggests that the patient may still have a calcium deficiency.

Yes. This confidence interval suggests that the patient no longer has a calcium deficiency.    

No. This confidence interval suggests that the patient may still have a calcium deficiency.

No. This confidence interval suggests that the patient no longer has a calcium deficiency.

Answer 1

	Values ( X )	Σ ( Xi- X̅ )2
	9.3	0.5929
	9	1.1449
	10.7	0.3969
	9.3	0.5929
	9.4	0.4489
	9.8	0.0729
	10	0.0049
	9.9	0.0289
	11.2	1.2769
	12.1	4.1209
Total	100.7	8.681

Part a)

Mean X̅ = Σ Xi / n
X̅ = 100.7 / 10 = 10.07

Sample Standard deviation S_X = √ ( (Xi - X̅ )² / n - 1 )
S_X = √ ( 8.681 / 10 -1 ) = 0.98

Part b)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.001 /2, 10- 1 ) = 4.781
10.07 ± t(0.001/2, 10 -1) * 0.98/√(10)
Lower Limit = 10.07 - t(0.001/2, 10 -1) 0.98/√(10)
Lower Limit = 8.5884 ≈ 8.59
Upper Limit = 10.07 + t(0.001/2, 10 -1) 0.98/√(10)
Upper Limit = 11.5516 ≈ 11.55
99.9% Confidence interval is ( 8.59 , 11.55 )

Part c)

No. This confidence interval suggests that the patient no longer has a calcium deficiency.