1) The world's smallest mammal is the bumblebee bat. The mean weight of 200 randomly selected bumblebee bats is 1.659 grams, with a standard deviation of 0.264 grams.
(a) Find a 99.9% confidence interval for the mean weight of all bumblebee bats at the following confidence levels (two places after decimal).
(b) Find a 99% confidence interval for the mean weight of all bumblebee bats at the following confidence levels (two places after decimal).
(c) Find a 95% confidence interval for the mean weight of all bumblebee bats at the following confidence levels (two places after decimal).
(d) Find an 80% confidence interval for the mean weight of all bumblebee bats at the following confidence levels (two places after decimal).
2) Dr. Clifford Jones claims that the mean weight of bumblebee bats is 1.7 grams. We are interested in whether it's less than he claims.
Using the t-distribution technique, is there evidence that bumblebee bats weigh less on average than he claims, at each of the following levels?

For  values

Hence the Dr.Clifford Jones claim is wrong about the bats mean weight

Step-by-step explanation:

Given:

Mean=1.659 grams

Standard deviation: 0.264

No of samples=200

To find :

Confidence intervals at

1)99.9% 2)99%  3)95% 4)80% and

Whether the Dr. Clifford with 1.7 mean weight is less or not?

Solution:

We know  interval estimation is given by ,

E=mean±Z value*{standard deviation/Sqrt(N)}

Now For Z value 99.9% =3.291

E=1.659±3.291{0.264/Sqrt(200)}

=1.659±0.0614

i.e.C.I.[1.6 to 1.72]

Now for Z value 99 % =2.576

E=1.659±2.576{0.264/Sqrt(200)}

=1.659±0.0481

i.e. C.I[1.61,1.71]

Now for Z value at 95% =1.96

E=1.659±1.96*(0.264/sqrt(200))

=1.659±0.0366

i.e. C.I.[1.62,1.7]

Now ofr Z value at 80% =1.28

E=1.659±1.28*(0.264/sqrt(20))

=1.659±0.0239

i.e. C.I.[1.64,1.68]

Using t distribution as ,

value for mean =1.7

raw i.p=1.659

Degree of freedom =N-1=200-1=199

Hence

t-score is similar to zscore

T-score =(Raw input -mean)/(standard deviation/Sqrt(n))

=(-1.7+1.659)/(0.264/Sqrt(200))

=-2.19721

Consider 1 tailed ,

p value =P(Z≤-2.19721)

=0.0143

i.e  P value is 0.0143

Hence The result is not significant at p<0.01

perimeter of the first rectangular board is 9 feet.

step-by-step explanation:

let p = perimeter of first board.

l = length of first board

w = width of first board

length of first board = 2 * width of first board - 3

l = 2w - 3

perimeter of first board = 2l + 2w

since l = 2w-3, then

p = 2*(2w-3) + 2w which equals 4w-6 + 2w which equals 6w-6

length and width of the second board are reciprocals of corresponding length and width of the first board.

length of second board = 1/(2w-3)

width of second board = 1/w

perimeter of second board is 1/5th perimeter of the first board.

perimeter of second board = p/5

perimeter of the second board = (2/(2w-3) + 2/w)

since the perimeter of the first board is 5 times the perimeter of the second board, this means that:

6w-6 = 5*(2/(2w-3) + 2/w)

which is the same as:

6w-6 = (10/(2w-3) + 10/w)

if we multiply both sides of this equation by (2w-3), we get:

(6w-6)*(2w-3) = 10 + 10*(2w-3)/w

if we multiply both sides of this equation by w, we get:

(6w-6)*(2w-3)*w = 10*w + 10*(2w-3)

we can simplify this to become:

(6w-6)*(2w-3)*w = 10w + 20w - 30

which becomes:

(6w-6)*(2w-3)*w = 30w-30

if we divide both sides of this equation by (6w-6), we get:

(2w-3)*w = (30w-30)/(6w-6)

which becomes:

(2w-3)*w = 5

this can be simplified to:

2w^2 - 3w = 5

subtract 5 from both sides of this equation to get:

2w^2 - 3w - 5 = 0

which can be factored into:

(2w-5)*(w+1) = 0

which makes:

w = 5/2

or

w = -1.

w can't be negative so the only possible answer is w = 5/2.

if w = 5/2, then

p = 2w-3 = 5-3 = 2

we have:

p = 2

w = 5/2

p = 2l + 2w gets p = 4 + 5 = 9

the perimeter of the first board is 9 feet.

since the perimeter of the second board is 1/5 the perimeter of the first board, the perimeter of the second board is 9/5.

to see if that's correct, we substitute known values for l and w into the second board.

the length of the second board is 1/l.

this becomes 1/2.

the width of the second board is 1/w

this bgecomes 1/(5/2) = (2/5)

let l2 = length of second board.

let w2 = width of second board.

l2 = (1/2)

w2 = (2/5)

let p2= perimeter of second board.

then p2 = 9/5

since p2 = 2*l2 + 2*w2, we should get p2 = 9/5 using the dimensions of l2 and w2.

2 * l2 = 2 * (1/2) = 1

2 * w2 = 2 * (2/5) = 4/5

1 is the same as 5/5

5/5 + 4/5 = 9/5

p2 checks out good.

the dimensions as given are good.

answer to the problem is:

perimeter of the first rectangular board is 9 feet.

ans789

step-by-step explanation: