Question

1.) An adult patient being treated for severe muscle spasms has her calcium level tested regularly because her condition could be attributed to a low calcium level. A random sample of her recent calcium levels (in mg/dL) is given below.

5.3, 6.8, 9.1, 8.9, 9.8,

5.9, 8.2, 6.1, 9.4, 9.0

(a) Find a 95% confidence interval for her mean calcium level.
(b) A person with a mean calcium level below 6 mg/dL is thought to have low calcium. Based upon your 95% confidence interval, does she have a low calcium level? Explain your reason(s).

Answer 1

The sample size is n = 10 . The provided sample data along with the data required to compute the sample mean and sample variance are shown in the table below:

	X	X2
	5.3	28.09
	6.8	46.24
	9.1	82.81
	8.9	79.21
	9.8	96.04
	5.9	34.81
	8.2	67.24
	6.1	37.21
	9.4	88.36
	9.0	81
Sum =	78.5	641.01

The sample mean is computed as follows:

Also, the sample variance is

Therefore, the sample standard deviation s is

(a) Find a 95% confidence interval for her mean calcium level.

The number of degrees of freedom are df = 10 - 1 = 9 and the significance level is α=0.05.

Based on the provided information, the critical t-value for α=0.05 and df = 9 degrees of freedom is t_c = 2.262

The 95% confidence for the population mean μ is computed using the following expression

Therefore, based on the information provided, the 95 % confidence for the population mean μ is

CI = (6.663, 9.037)

which completes the calculation.

(b) A person with a mean calcium level below 6 mg/dL is thought to have low calcium. Based upon your 95% confidence interval, does she have a low calcium level? Explain your reason(s).

The confidence interval is given by CI = (6.663, 9.037)

And the interval does not include the value 6 hence the level is low