A machine used for packaging seedless golden raisins is set so that their standard deviation in the weight of raisins packaged per box is 0.25 ounce. The operations manager wishes to test the machine setting and selects a sample of 30 consecutive raisin packages filled during the production process. Weights
15.2
15
14.3
15.3
15.2
14.4
15.1
15.4
15.5
15.7
15.6
15.4
15.3
15.7
15.2
15
15.4
15.5
15.1
15.3
15.6
14.3
14.9
15.1
14.6
14.8
15.3
14.5
14.6
15.1
At the 0.05 level of significance, is there evidence that the population standard deviation differs from 0.25 ounces?
What assumptions are made in order to perform this test?

population standard deviation does not differ from 0.25

Step-by-step explanation:

The hypothesis :

σ² = 0.25²

σ² ≠ 0.25²

The Chisquare statistic, χ²: (n-1)*s²/σ²

Where s² = sample variance ; sample size, n = 30

Using calculator, the sample variance, s² = 0.1646

χ² = (n-1)*s²/σ²

χ²: (30 - 1) * 0.1646 / 0.25²

χ² = (29 * 0.1646) / 0.0625

χ² = 0.3292 / 0.0625

χ² = 5.2672

α = 0.05

Using the Pvalue from Chisquare calculator ; df = 29

Pvalue = 1

Since, Pvalue > α ; We fail to reject the null and conclude that population standard deviation does not differ from 0.25

just do 5.2 + 2 ( 7.2) and 1/3 x 3 (.9) and youv'e got your equation. 0.9( 9/10) + 7.2 ^2 = 16.4 hope i could !

step-by-step explanation:

the correct answer is c