STA301 Assignment 4 Solution

Assignment # 04

STA 301 (Statistics and probability)

Question # 01

a)

Answer:

            To give our Sampling distribution a state of normalization .In other words, As the sample size approaches infinity our sampling distribution of mean of X approaches normality  Due to Central Limit Theorem, the normal distribution has found a central place in statistical inference

b)

    “An estimate is a numerical value of the unknown parameter obtained by applying a rule or formula called an “estimator to the sample” taken from a population for instance, if X1, X2, X3…….Xn is a random sample of size n from a population with mean µ then,

                         i

X = 1/n ∑ Xi   is called its estimator

                        n =1

Unknown parameter could be mean, median, mode, proportion or standard deviation In general, it is denoted by θ and its estimator is denoted by T

 c) For any population parameter θ and its estimator T, the quantity

           E (T) – θ   is known as amount of bias

If this quantity is equal to zero then we say that our estimator is unbiased since unbiased ness is a desirable quality so we can regard our estimator as good one

 d) The expression for confidence interval is given by

  

                                     X + Z ά/2 σ/√n

For example in case of mean     

There are two situations to achieve precision

1) Increase the level of significance or 

2)  Making the “n” small

By fulfilling one of these conditions the other becomes less précised so we need to strike a compromise between how low a level of confidence we can achieve and how wide an interval we can tolerate

e)  27.5 <   < 43.8

   Means that it is estimated that population mean lies somewhere between 27.5 and 43.8 and this statement is made on the basis of 85% confidence

Question # 02

a)

b)                              e   = 0.05

                                 q   = 1 – p = 0.84

                                   p = 0.16

                         Z 0.025 = 1.96 

Formula for determining value of sample size in case of population proportion

                                    n   = [Z ά/2] pq      

                                                 e 

Putting values

                        n = [1.96 / 0.05] ^ 2 x (0.16) (0.84)

                        n = 206.52

                        n = 207