Answer to Question #163879 in Statistics and Probability for ayesha

Question

Approximately 60% of all part-time college students in the Peshawar are female. What would

you expect to see in terms of the behavior of a sample proportion of females if random samples of

size 100 were taken from the population of all part-time college students? Estimate the standard

deviation also construct the confidence interval using alpha 0.05.

Answer 1

As we know, due to sampling variability, sample proportion in random samples of size $100$

 will take numerical values which vary according to the laws of chance:

In other words, sample proportion is a random variable. To summarize the behavior of any random variable, we focus on three features of its distribution: the center, the spread, and the shape.

Based only on our intuition, we would expect the following:

Center: Some sample proportions will be on the low side — say, $0.55 \text{or} 0.58$ — while others will be on the high side — say, $0.61 \text{or} 0.66$ . It is reasonable to expect all the sample proportions in repeated random samples to average out to the underlying population proportion,

 $0.6$ . In other words, the mean of the distribution of p-hat should be p.

Spread: For samples of $100$ , we would expect sample proportions of females not to stray too far from the population proportion $0.6$ .Sample proportions lower than $0.5$ or higher than $0.7$ would be rather surprising.

On the other hand, if we were only taking samples of size $10$ , we would not be at all surprised by a sample proportion of females even as low as $\dfrac{4}{10} = 0.4,$ or as high as $\dfrac{8}{10} = 0.8.$ Thus, sample size plays a role in the spread of the distribution of sample proportion:there should be less spread for larger samples, more spread for smaller samples.

Shape: Sample proportions closest to $0.6$ would be most common, and sample proportions far from $0.6$ in either direction would be progressively less likely. In other words, the shape of the distribution of sample proportion should bulge in the middle and taper at the ends: it should be somewhat normal.

The standard deviation is given by the formula-

$\sigma_p=\sqrt{\dfrac{p(1-p)}{N}}$

$=\sqrt{\dfrac{0.6(1-0.6)}{100}}=\sqrt{\dfrac{0.24}{100}}$ =$\dfrac{0.489}{10}=0.0489$

So The two confidence interval for $\alpha=0.05 \text{ is given by- }$

The probability is approximately 0.95 that p-hat falls within 2 standard deviations of the mean

=, $0.6-2(\alpha)=0.6 – 2(0.05) =0.5\text{ and } 0.6+2(\alpha)= 0.6 + 2(0.05)=0.7.$

Hence, TheConfidence interval is (0.5,0.7)