Question

The fracture strengths of a certain type of glass have a average  μ and a standard deviation 2. A random sample of size 100 from this glass has a sample mean 15.5

a. Find a 95% confidence interval for the average fracture strength.

b. If we assume that μ = 15, what is the probability that the average of fracture strength for 100 randomly selected pieces of this glass exceeds 15.5?

c. Is there enough evidence to conclude that the fracture strength average is more than 15?. Use the level of significance α = .05.

d. Do you use t-test or z-test? Why?

Answer 1

a)

sample mean, xbar = 15.5
sample standard deviation, σ = 2
sample size, n = 100

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

ME = zc * σ/sqrt(n)
ME = 1.96 * 2/sqrt(100)
ME = 0.39

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (15.5 - 1.96 * 2/sqrt(100) , 15.5 + 1.96 * 2/sqrt(100))
CI = (15.108 , 15.892)

b)

Here, μ = 15, σ = 0.2 and x = 15.5. We need to compute P(X >= 15.5). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (15.5 - 15)/0.2 = 2.5

Therefore,
P(X >= 15.5) = P(z <= (15.5 - 15)/0.2)
= P(z >= 2.5)
= 1 - 0.9938 = 0.0062

c)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 15
Alternative Hypothesis, Ha: μ > 15

Rejection Region
This is right tailed test, for α = 0.05
Critical value of z is 1.645.
Hence reject H0 if z > 1.645

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (15.5 - 15)/(2/sqrt(100))
z = 2.5

P-value Approach
P-value = 0.0062
As P-value < 0.05, reject the null hypothesis.

d)

z test because sigma is known with large sample size