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 .

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

step-by-step explanation:

milk=16 large

soda=280 kids

juice box=82 regular

63 soda +63 regular

124 large total

14

step-by-step explanation: