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

8.4

10.3

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 =	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:

Given that,

x	x2
9.3	86.49
8.4	70.56
10.3	106.09
8.5	72.25
9.4	88.36
9.8	96.04
10	100
9.9	98.01
11.2	125.44
12.1	145.41
x = 98.9	x2 = 989.65

Solution:

Given that

a ) The sample mean is

Mean   = (x / n)

= ( 9.3+ 8.4+10.3+8.5+9.4+9.8+10.0+9.9+11.2+12.1 / 10 )

= 98.9 / 10

= 9.89

x = 9.89 mg / dl

The sample standard is S

  S = ( x ² ) - (( x)² / n ) / 1 -n )

= ( 989.65( - ( 98.9 )² / 10 ) / 9

   = (989.65 978.121 / 9 )

= 11.529 / 9

= 1.281

= 1.1318

The sample standard s  = 1.13 mg / dl

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 = 0.0005

t _/2,df = t_0.0005,9 =2.797

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

= 4.781 * (1.13 / 10)

= 1.71

Margin of error = 1.71

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

- E < < + E

9.89 - 1.71 < < 9.89 + 1.71

8.18 < < 11.6

Lower limit = 8.18 mg /dl

Upper limit = 11.6 mg /dl