A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 417 gram setting. Based on a 26 bag sample where the mean is 421 grams and the standard deviation is 15, is there sufficient evidence at the 0.05 level that the bags are underfilled or overfilled? Assume the population distribution is approximately normal. State the null and alternative hypotheses.

Null, H0 : μ = 417

Alternative, H1 : μ ≠ 417

Step-by-step explanation:

Given that :

The mean grams by a bag filling machine μ = 417

Hence, the null hypothesis is That bag filling machine works correctly at 417 grams setting

The alternative hypothesis will negate the null ; In this question, the alternative hypothesis looks at the possibility of either overfill or underfill, which In straightforward terms means that the 417 g ram setting for the bag filling machine is incorrect.

Hence,

Null, H0 : μ = 417

Alternative, H1 : μ ≠ 417

add 5 subtract 1

step-by-step explanation:

answer: ratio of volume of cone (with 3 times the height and equal diameter) to volume of cylinder is 2 to 1. the value of π is not used.

step-by-step explanation:

radius is diameter/2 = 4.

area of circle is π r^2.

volume of cylinder is height × area of base = hπr^2 = 3(4^2) π = 48π cubic inches.

not knowing the formula for volume of cone, i calculate it using calculus (which is for calculating things).

volume of cone is integral 0 to h area(y) dy. area(y) is π r(y)^2. r(y) runs r to zero as y runs zero to h, r(0) = r, r(h) = 0.

r(y) = r - r/h y

area(y) = π (r - r/h y)^2 = π (r^2 - 2r^2y/h + r^2y^2/h^2) = π r^2(1-2y/h+y^2/h^2)

antiderivative area(y) = π r^2 (y-y^2/h+y^3/(3h^2))

integral 0 to h = π r^2(h-h^2/h+h^3/(3h^2) - (0)) = π r^2(h - h + h/3) = π r^2(h/3) = π/3 r^2 h

volume of this cone is π/3(4^2)18 cubic inches.

= 6 16 π = 96π cubic inches.

cone has twice the volume, 96/48.

cone height is 6 times the cylinder.

ratio of volume formula: π r^2 cancels.

cone over cylinder ratio is 6h/3 to h, 2 to 1.