Question

1. The temperature at which metallic glass becomes brittle was of interest. Now suppose we would like to determine whether this mean temperature exceeds 595⁰ C. A random sample of 20 is taken yielding a mean of 600⁰ C and sample standard deviation of 15⁰ C. (Use t-distribution)

1. Find the 95% confidence interval for the mean
2. Test the hypothesis at a level of significance of alpha = 0.05. What is your conclusion?

Answer 1

a)
sample mean, xbar = 6000
sample standard deviation, s = 150
sample size, n = 20
degrees of freedom, df = n - 1 = 19

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.093

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (6000 - 2.093 * 150/sqrt(20) , 6000 + 2.093 * 150/sqrt(20))
CI = (5929.8 , 6070.2)

b)
Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 5950
Alternative Hypothesis, Ha: μ > 5950

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (6000 - 5950)/(150/sqrt(20))
t = 1.491

P-value Approach
P-value = 0.0762
As P-value >= 0.05, fail to reject null hypothesis.

There is not sufficient evidence to conclude that mean temperature exceeds 5950 C