In two common species of flowers, A and B, the proportions of flowers that are blue are and , respectively. Suppose that independent random samples of 50 species-A flowers and 100 species-B flowers
are selected. Let be the sample proportion of blue species-A flowers and be the sample proportion of
blue species-B flowers. What is the mean of the sampling distribution of ?

A) Pa- Pb
B)
C) Pa^ - PB^
D)
E) [tex]\sqrt{\frac{Pa(1-pa)}{50}+ [tex]

The mean of the sampling distribution of pₐ^ - (p_b)^ is;

Option A; Pa - Pb

We are given;

Number of independent samples of species A flowers; n_a = 50

Number of independent samples of species B flowers; n_b = 100

We are told that;

pₐ^ is the sample proportion of samples of species A flowers

(p_b)^ is the sample proportion of samples of species A flowers

Now, we want to find the mean of the sampling distribution of pₐ^ - (p_b)^.

Formula for the mean here is;

p^ = x/n

Furthermore, it is stated in formula for mean of sample of distribution is;

E(p^) = μ_p = p

Thus; E(p^) = p

Then;

E( pₐ^ ) = pa

Similarly; E(p_b)^ = pb

Thus;

E(pₐ^ - (p_b)^) = pa - pb

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A.) Pa - Pb

Explanation:

Given that:

A and B are common species of flower ;

Proportion of blue flowers are Pa and Pb for the two species respectively ;

Sample Proportion of blue flower;

Species A = P'a

Species B = P'b

Mean of sampling distribution P'a - P'b

Since the samples are independent and drawn at random, the mean of sampling distribution equals Pa - Pb