Suppose the level of calcium in the blood in healthy young a

Suppose the level of calcium in the blood in healthy young adults varies with mean about 9.4 milligrams per deciliter and standard deviation about ? = 0.3. A clinic in rural Guatemala measures the blood calcium level of 140 healthy pregnant women at their first visit for prenatal care. The mean is

x = 9.46.

Is this an indication that the mean calcium level in the population from which these women come differs from 9.4?

(a) State H₀ and H_a.

H₀: ? < 9.4 mg/dl; H_a: ? = 9.4 mg/dl

H₀: ? > 9.4 mg/dl; H_a: ? = 9.4 mg/dl

H₀: ? = 9.4 mg/dl; H_a: ? > 9.4 mg/dl

H₀: ? = 9.4 mg/dl; H_a: ? ? 9.4 mg/dl

H₀: ? = 9.4 mg/dl; H_a: ? < 9.4 mg/dl

(b) Carry out the test and give the P-value, assuming that ? = 0.3 in this population. (Round your answer to four decimal places.)


Report your conclusion.

There is sufficient evidence at the 5% level that ? is different from 9.4 mg/dl. There is not sufficient evidence at the 5% level that ? is less than 9.4 mg/dl.     There is sufficient evidence at the 5% level that ? is greater than 9.4 mg/dl. There is not sufficient evidence at the 5% level that ? is different from 9.4 mg/dl. There is sufficient evidence at the 5% level that ? is less than 9.4 mg/dl.

(c) Give a 95% confidence interval for the mean calcium level ? in this population. We are confident that ? lies quite close to 9.4. This illustrates the fact that a test based on a large sample (n = 140 here) will often declare even a small deviation from H₀ to be statistically significant. (Round your answers to three decimal places.)
( , )

Solution

(a)H₀: ? = 9.4 mg/dl; H_a: ? ? 9.4 mg/dl

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(b) The test statistic is

Z=(xbar-mu)/(s/vn)

=(9.46-9.4)/(0.3/sqrt(140))

=2.37

It is a two-tailed test.

So the p-value= 2*P(Z>2.37) =0.0178 (from standard normal table)

There is sufficient evidence at the 5% level that ? is different from 9.4 mg/dl.

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(c) Given a=1-0.95=0.05, Z(0.025) = 1.96 (from standard normal table)

So the lower bound is

xbar - Z*s/vn =9.46 -1.96*(0.3/sqrt(140)) =9.410

So the upper bound is

xbar + Z*s/vn =9.46 +1.96*(0.3/sqrt(140)) =9.510