Question

The producer of a certain bottling equipment claims that the variance of all its filled bottles is .027 or less. A sample of 30 bottles showed a standard deviation of .2. The p-value for the test is _____.

Answer 1

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

Answer 2

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.