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.
Values ( X ) | Σ ( X_{i}- 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.
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