Question

1.(10) Assume that the proportion of successes in a population is p. If simple random samples of size n are drawn from the population and the proportions, p. of successes in the samples are calculated, then the distribution of the sample proportions p is normal. What are the mean and standard deviation of this Normal distribution? Hp = 2.(10) How large do the number of successes and the number of failures in a sample have to be in order to use a large sample confidence interval? 3.(20) In order to calculate the size of a sample necessary for the margin of error, m, to be some specific number one needs to make a guess at a proportion, p*, to use in calculating the standard error. What is the most conservative guess for p*? Use the most conservative guess to determine the smallest sample size that will guarantee a margin of error of .02 for a 90% confidence interval.

Answer 1

1) $\mu_{\widehat p} = p$

   $\sigma_{\widehat p} = \sqrt{\frac{p(1 - p)}{n}}$

2) The number of success and the number of failures should have 10 in order use a large sample confidence interval.

3) p^* = 0.5

At 90% confidence level, the critical value is z_0.05 = 1.645

Margin of error = 0.02

$or, z_{0.05} * \sqrt{\frac{p(1 - p)}{n}} = 0.02$

$or, 1.645 * \sqrt{\frac{0.5(1 - 0.5)}{n}} = 0.02$

$or, n = (1.645 * \sqrt{\frac{0.5(1 - 0.5)}{0.02}})^2$

4) H₀: P < 0.5

    H_a: P > 0.5

$\widehat p$ = 61/100 = 0.61

$z = \frac{\widehat p - P}{\sqrt{\frac{P(1 - P)}{n}}}$

    $= \frac{0.61 - 0.5}{\sqrt{\frac{0.5(1 - 0.5)}{100}}}$

   

P-value = P(Z > 2.2)

             = 1 - P(Z < 2.2)

             = 1 - 0.9861

             = 0.0139

Since the P-value is less than the significance level(0.0139 < 0.05), so we should reject the null hypothesis.

At 0.05 significance level, there is sufficient evidence to conclude that a majority of seniors think their job prospects are good.

5) The 95% confidence interval is

$\widehat p \pm z_{0.025} * \sqrt{\frac{p(1 - p)}{n}}$

$= 0.61 \pm 1.96 * \sqrt{\frac{0.61(1 - 0.61)}{100}}$

$= 0.61 \pm 0.0956$