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 10.9 8.9 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 bar and the sample standard deviation s. (in mg/dl; round your answers to two decimal places.) x bar = mg/dl s = mg/dl

(b) Find a 99.9% confidence interval for the population mean of total calcium in this patient's blood. (in mg/dl; round your answer to two decimal places.) lower limit mg/dl upper limit mg/dl

Answer 1

Solution :

We are given a data of sample size n = 10

9.5,8.8,10.9,8.9,9.4,9.8,10.0,9.9,11.2,12.1

Using this, first we find sample mean() and sample standard deviation(s).

=   

= (9.5 + 8.8.......+ 12.1)/10

= 10.05

Now ,

s =   

s = (9.5 - 10.05)2 + (8.8 - 10.05)2 + ........ + (12.1 - 10.05)2  / 10 - 1

s = 1.05

Using given data, find Xi - for each term.Take square for each.Then we can easily find s.

s = 1.05

Note that, Population standard deviation() is unknown..So we use t distribution.

Our aim is to construct 99.9% confidence interval.

   c = 0.999

= 1 - c = 1 - 0.999 = 0.001

  /2 = 0.001 2 = 0.0005

Also, d.f = n - 1 = 9  

    =    =  0.0005,9 = 4.781

( use t table or t calculator to find this value..)

The margin of error is given by

E =  /2,d.f. * ( / n )

= 4.781* ( 1.05/ 10 )

= 1.587

Now , confidence interval for mean() is given by:

( - E ) <   <  ( + E)

( 10.05 - 1.587 )   <   <  ( 10.05 + 1.587 )

8.46 <   < 11.64

Required 99.9% confidence interval is ( 8.46 , 11.64 )

The sample mean = 10.05 mg/dl

The sample standard deviation s = 1.05 mg/dl

Lower limit = 8.46 mg/dl

Upper limit = 11.64 mg/dl