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. 9.5 8.8 10.5 8.5 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 = 9.97 mg/dl s = 1.08 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 8.34 mg/dl upper limit 11.60 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

Given

Assume that the population of x values has an approximately normal distribution.

Data :- 9.5 8.8 10.5 8.5 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.

Formula :-

Sample mean = = ( 9.5 +8.8+ 10.5+ 8.5 +9.4 +9.8 +10.0 +9.9 +11.2 +12.1 ) / 10

                                       = 99.7 / 10 = 9.97

Sample mean = 9.97

Sample standard deviation s =

s =

s = 1.077085 1.08                     { after calculation }

Thus

Sample mean = 9.97 mg/dl

Sample standard deviation s = 1.08 mg/dl

(b) Find a 99.9% confidence interval for the population mean of total calcium in this patient's blood.

99% confidence interval is given by

CI = { - * , + *    }

Here is t-distributed with n-1 = 10-1 = 9 degree of freedom (DF) with = 0.001 { for 99.9% confidence }

It can be calculated from statistical book or from any software like R/Excel

From R

> qt(1-0.001 / 2 , df=9 )
[1] 4.780913

Thus = 4.780913

Hence confidence interval will be

CI = { - * , + *    }

    = { 9.97 - 4.780913 * , 9.97 + 4.780913 *    }

    = { 8.337194   , 11.60281 }

Thus LL = 8.337194    8.34 and    UL = 11.60281 11.60

Thus , 99.9% confidence interval for the population mean of total calcium in this patient's blood is { 8.34 ,11.60}

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

Explain :-

Now , we are given that , an adult patient has been treated for tetany if an average total calcium level below 6 mg/dl.

Since our 99.9% confidence interval has both Lower Limit LL = 8.34 and Upper Limit UL = 11.60 both greater that 6 , so with 99.9% confidence we can say that we are 99.9% sure that this patient do not has any calcium deficiency .

hence Correct Option is

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

Option 4) No. This confidence interval suggests that the patient no longer has a calcium deficiency.