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.
