A manufacturer wants to inspect the absorption capacity of a new sponge. Based on past data, the average sponge could absorb 3.5 ounces. After the redesign, the absorption amounts of a sample of 10 sponges were (in ounces): 4.1, 3.7, 3.3, 3.5, 3.8, 3.9, 3.6, 3.8, 4.0, and 3.9. At a level of significance of 1%, would you claim that the new design is better than the old one? Requried:
a. No because we don't reject the Null Hypothesis at Alpha = 0.01
b. Yes because we reject the Null hypothesis at Type 1 error = 1%
c. Yes because the cut-off is more than the sample mean
d. No because the cut-off is less than the sample mean

c. Yes because the cut-off is more than the sample mean

Step-by-step explanation:

The null and alternate hypotheses are

H0 : u= 3.5 ounces  vs Ha: u > 3.5 ounces

The mean and standard deviation of the sample are calculated to be 3.76 and 0.24

The test statistics is

t= x`- u/s√n

= 3.76-3.5/0.24/√10

= 0.26/0.07589

=3.426

The critical region for z test is  z > ± 2.33

The critical region for t test is   t > ± 2.82 for 9 d.f

For both the tests( z and t)   the test statistic =3.426 lies  in the critical region.

Hence we reject H0 and accept the hypothesis that  the new design is better than the old one at a level of significance of 1%.

cut-off = mean + 3 times the standard deviation

          =  3.76 + 3* 0.24

 cut-off     =4.48

Yes, the cut off is more than the sample mean.

So choice c is the best option.

3 is greater then both of those numbers

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.