Answer:
The p-value for the test is 0.0459.
Explanation:
The question involves a chi-squared test whose p-value is to be determined.
H₀: σ² ≤ 0.027 (null hypothesis)
H₁: σ² > 0.027 (alternative hypothesis)
Standard deviation = s = 0.2
Hence, s² = (0.2)² = 0.04
Sample size = n = 30
Degree of freedom = n - 1 = 30 - 1 = 29
Significance level = 0.05
Test statistic: X² = (n - 1)s² / σ²
= (30 - 1) x 0.04 / 0.027
= 42.9629
The p-value can now be determined using the Excel function:
CHISQ.DIST.RT(42.9629,29) = 0.0459
Hence, the p-value for the test is 0.0459.
Answer:
p-value = 0.0458
Step-by-step explanation:
We are given sample size; n = 30
Variance = 0.027
Standard deviation; S = 0.2
The value we want to test will be given by;
σ_o = √var = √0.027 = 0.1643
α = significance level
Before we test the hypothesis, let's calculate the statistic given by;
t = (n-1)[s/σ_o]²
So, t = (30 - 1)[0.2/0.1643]² = 42.972
Now, let's state the hypothesis ;
H1; σ ≤ 0.027
H0; σ > 0.027
The statistic has a Chi Square distribution distribution with degree of freedom as (n - 1). Thus,
The degree of freedom is,
n - 1 = 30 - 1 = 29
Using a chi square calculator online, P-value with chi square of 42.972 and degree of freedom as 29, p = 0.0458
