Question

Here are estimates of the daily intakes of calcium (in mg) for a sample of women between the ages of 18 and 24 who participated in a national study of women's bone health.

data

808

882

1062

970

909

802

374

416

784

997

651

716

438

1420

1425

948

1050

976

572

403

626

774

253

549

1325

446

465

1269

671

696

1156

684

1933

748

1203

2433

1255

1100

a) Find a 95% confidence interval for the mean.

b) Suppose that the recommended daily allowance (RDA) of calcium for women in this age range is 1200mg. State an appropriate null and alternative hypothesis, give the test statistic, the P-value, and state your conclusion.

Answer 1

a)

sample mean, xbar = 899.7105
sample standard deviation, s = 437.226
sample size, n = 38
degrees of freedom, df = n - 1 = 37

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.026

ME = tc * s/sqrt(n)
ME = 2.026 * 437.226/sqrt(38)
ME = 143.7

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (899.7105 - 2.026 * 437.226/sqrt(38) , 899.7105 + 2.026 * 437.226/sqrt(38))
CI = (756.01 , 1043.41)

b)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 1200
Alternative Hypothesis, Ha: μ ≠ 1200

Rejection Region
This is two tailed test, for α = 0.05 and df = 37
Critical value of t are -2.026 and 2.026.
Hence reject H0 if t < -2.026 or t > 2.026

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (899.7105 - 1200)/(437.226/sqrt(38))
t = -4.234

P-value Approach
P-value = 0.0001
As P-value < 0.05, reject the null hypothesis.