Question

Listed below is are lead concentrations in some medicines. Use a 0.05 significance level to test the claim that the mean lead concentration is less than 14. x̄=11.05 Sx= 6.46

3.0, 6.5, 6.0, 5.5, 20.5, 7.5, 12.0, 20.5, 11.5, 17.5

Critical Values: z0.005 = 2.575, z0.01 = 2.325, z0.025 = 1.96, z0.05 = 1.645

When d.f.=6: t0.005 = 3.707, t0.01 = 3.143, t0.025 = 2.447, t0.05 = 1.943

When d.f.=9: t0.005 = 3.250, t0.01 = 2.821, t0.025 = 2.262, t0.05 = 1.833

When d.f.=44: t0.005 = 2.690, t0.01 = 2.412, t0.025 = 2.014, t0.05 = 1.679

When d.f.=100: t0.005 = 2.625, t0.01 = 2.364, t0.025 = 1.984, t0.05 = 1.660

Answer 1

Here,sample standard deviation is given so we use t test

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 14
Alternative Hypothesis, Ha: μ < 14

Rejection Region
This is left tailed test, for α = 0.05 and df = 9
Critical value of t is -1.833.
Hence reject H0 if t < -1.833

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (11.05 - 14)/(6.46/sqrt(10))
t = -1.444

P-value Approach
P-value = 0.0913
As P-value >= 0.05, fail to reject null hypothesis.