Question

A) a sample of 101 body temperatures of randomly selected people has a mean of 98.20 degrees Fahrenheit and a standard deviation of 0.62 degrees Fahrenheit. Construct a 95% confidence level confidence interval for the population mean body temperature.

B) Use the above result to test the claim that the population mean body temperature is 98.6 degrees Fahrenheit with significance level 0.05.

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

a)

sample mean, xbar = 98.2
sample standard deviation, s = 0.62
sample size, n = 101
degrees of freedom, df = n - 1 = 100

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 1.984

ME = tc * s/sqrt(n)
ME = 1.984 * 0.62/sqrt(101)
ME = 0.1

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (98.2 - 1.984 * 0.62/sqrt(101) , 98.2 + 1.984 * 0.62/sqrt(101))
CI = (98.0776 , 98.3224)

b)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 98.6
Alternative Hypothesis, Ha: μ ≠ 98.6

Rejection Region
This is two tailed test, for α = 0.05 and df = 100
Critical value of t are -1.984 and 1.984.
Hence reject H0 if t < -1.984 or t > 1.984

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (98.2 - 98.6)/(0.62/sqrt(101))
t = -6.484

P-value Approach
P-value = 0
As P-value < 0.05, reject the null hypothesis.