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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?

Respuesta :

Answer:

For  values

  1. 99.9%==1.659±0.0614
  2. 99%==1.659±0.0481
  3. 90%==1.659±0.0366
  4. 80%=1.659±0.0239

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

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