Question

Let x be a random variable that represents hemoglobin count (HC) in grams per 100 milliliters of whole blood. Then x has a distribution that is approximately normal, with population mean of about 14 for healthy adult women. Suppose that a female patient has taken 10 laboratory blood tests during the past year. The HC data sent to the patient's doctor are as follows.

14

18

17

18

14

13

15

16

15

11

(i) Use a calculator with sample mean and standard deviation keys to find x and s. (Round your answers to two decimal places.)

x	=
s	=

(ii) Does this information indicate that the population average HC for this patient is higher than 14? Use α = 0.01.

(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: μ = 14; H₁: μ ≠ 14H₀: μ = 14; H₁: μ < 14     H₀: μ < 14; H₁: μ = 14H₀: μ = 14; H₁: μ > 14H₀: μ > 14; H₁: μ = 14

(b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution.

The Student's t, since we assume that x has a normal distribution and σ is unknown.The standard normal, since we assume that x has a normal distribution and σ is unknown.    The Student's t, since we assume that x has a normal distribution and σ is known.The standard normal, since we assume that x has a normal distribution and σ is known.

What is the value of the sample test statistic? (Round your answer to three decimal places.)


(c) Estimate the P-value.

P-value > 0.2500.100 < P-value < 0.250    0.050 < P-value < 0.1000.010 < P-value < 0.050P-value < 0.010

Sketch the sampling distribution and show the area corresponding to the P-value.

Answer 1

We will first make certain computations here as:

x	X - Mean(X)	(X - Mean(x))^2
14	-1.1	1.21
18	2.9	8.41
17	1.9	3.61
18	2.9	8.41
14	-1.1	1.21
13	-2.1	4.41
15	-0.1	0.01
16	0.9	0.81
15	-0.1	0.01
11	-4.1	16.81
151	0	44.9

'The last line shows the sum of the given column.

i) The sample mean and sample standard deviation here are computed as:

The sample standard deviation now is computed here as:

Therefore 2.2336 is the required sample standard deviation here.

ii) a) We are given the level of significance as 0.01 that is given in the problem.

As we are testing here whether the mean is greater than 14, therefore the null and the alternate hypothesis here are given as:

b) As we are not given the population standard deviation here and the sample size is also very small, therefore the t distribution is to be used here. The Student's t, since we assume that x has a normal distribution and σ is unknown

The test statistic here is computed as:

Therefore 1.557 is the test statistic value here.

c) For n - 1 = 9 degrees of freedom, we get from the t distribution tables:

p = P(t9 > 1.5574) = 0.06

Therefore the p-value lies in the range 0.05 < p < 0.1

This is shown in the graph form as: