The desired percentage of sio2 in a certain type of aluminous cement is 5.5. to test whether the true average percentage is 5.5 for a particular production facility, 16 independently obtained samples are analyzed. suppose that the percentage of sio2 in a sample is normally distributed with σ = 0.32 and that x = 5.24. (use α = 0.05.) (a) does this indicate conclusively that the true average percentage differs from 5.5?
state the appropriate null and alternative hypotheses.

h0: μ = 5.5
ha: μ > 5.5h0: μ = 5.5
ha: μ ≥ 5.5 h0: μ = 5.5
ha: μ < 5.5h0: μ = 5.5
ha: μ ≠ 5.5

z=-3.25

Step-by-step explanation:  

1) Data given and notation  

represent the mean production for the sample  

represent the sample standard deviation for the sample  

sample size  

represent the value that we want to test  

represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

represent the p value for the test (variable of interest)  

Part a: State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean production is different from 5.5 tons, the system of hypothesis would be:  

Null hypothesis:  

Alternative hypothesis:  

If we analyze the info given we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

(1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

P-value

Since this is a two side test the p value would be:  

Conclusion  

If we compare the p value and the significance level given we see that so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the production its significant different compared to the desired percentage of SiO2 of 5.5 at 5% of signficance.  

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.

step-by-step explanation:

i found the koficent of the linear function and i foun c acordind to the strict formula y=kx+c and then i took 2 as an x and replacet it on the formula i found and then foun the y