Question

33) 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

10.5

9.1

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

Answer 1

Solution:

x	x2
9.3	86.49
9	81
10.5	110.25
9.1	82.81
9.4	88.36
9.8	96.04
10	100
9.9	98.01
11.2	125.44
12.1	146.41
x=100.3	x2=1014.81

a )The sample mean is

Mean     = (x / n) )

= (9.3+ 9.0 +10.5 +9.1 + 9.4 + 9.8 + 10.0 + 9.9 + 11.2 + 12.1/ 10 )

= 100.3 / 10

= 10.03

Mean    = 10.03

The sample standard is S

  S = ( x² ) - (( x)² / n ) n -1

= (1014.81( (100.3)² / 10 ) 9

   = ( 1014.81- 1006.009 / 9)

= (8.801/ 9)

= 0.9779

= 0.9889

The sample standard is = 0.99

b ) Degrees of freedom = df = n - 1 = 10 - 1 = 9

At 99.9% confidence level the t is ,

= 1 - 99.9% = 1 - 0.999 = 0.001

/ 2 = 0.001 / 2 = 0.005

t _/2,df = t_0.0005,9 =4.781

Margin of error = E = t_/2,df * (s /n)

= 4.781 * (0.99/ 10)

= 2.24

Margin of error = 2.24

The 99.9% confidence interval estimate of the population mean is,

- E < < + E

10.03 - 2.24< < 10.03 + 2.24

7.79 < < 12.27

Lower limit = 7.79

Upper limit = 12.27