Question

the first drop down menu is increase/decrease and the second is larger/smaller.

eBook According to Reader's Digest, 49% of primary care doctors think their patients receive unnecessary medical care. a. Suppose a sample of 330 primary care doctors was taken. Use z-table. Show the sampling distribution of the proportion of the doctors who think their patients receive unnecessary medical care. ơi-| J(to 4 decimals) t is the probability that the sample proportion will be within t0.08 of the population proportion. Round your answer to four declmals. c. What is the probability that the sample proportion will be within s0.06 of the population pr opertion. Round your answer to four decimals d. What would be the effect of taking a larger sample on the probabilities in parts (b) and (c)? Why? しSelect your answer .. This is because the increase in the sample size makes the standard er ror, op Select your answr

Answer 1

The proportion of the doctors who think their patients receive unnecessary care is p=0.49 (according to Reader's Digest, and hence we will treat this as population proportion).

a) A sample of size n=330 is taken.

The expected value of the sample proportion is

$E(\bar{p})=p=0.49$

The standard error of proportion is

$\sigma_{\bar{p}}=\sqrt\frac{p(1-p)}{n}=\sqrt\frac{0.49(1-0.49)}{330}=0.0275$

ans: $E(\bar{p})=0.49$

$\sigma_{\bar{p}}=0.0275$

Since $np=330\times 0.49=161.7\text{ and }n(1-p)=330\times (1-0.49)=168.3$ are both greater than 5,

we can use normal distribution to approximate the sampling distribution of proportions.

Let $\bar{P}$ be the sample proportion of doctors who think their patients receive unnecessary care, for a sample of size n=330.

$\bar{P}$ has a normal distribution with mean $\mu_{\bar{P}}=0.49$ and standard error $\sigma_{\bar{P}}=0.0275$

b) The probability that the sample proportion will be within +-0.03 of the population proportion (which is 0.49) is given by the probability that $\bar{P}$ is between 0.49-0.03=0.46 and 0.49+0.03=0.52

$\begin{align*} P(0.46<\bar{P}<0.52)&=P\left(\frac{0.46-\mu_{\bar{p}}}{\sigma_{\bar{p}}} <\frac{\bar{P}-\mu_{\bar{p}}}{\sigma_{\bar{p}}}<\frac{0.52-\mu_{\bar{p}}}{\sigma_{\bar{p}}}\right )\quad\text{get the z-scores of 0.46 and 0.52}\\ &=P\left(\frac{0.46-0.49}{0.0275} <Z<\frac{0.52-0.49}{0.0275}\right )\\ &=P(-1.09<Z<1.09)\\ &=P(Z<1.09)-P(Z<-1.09)\\ &=P(Z<1.09)-P(Z>1.09)\\ &=P(Z<1.09)-(1-P(Z<1.09))\\ &=2P(Z<1.09)-1\\ &=2\times (0.5&plus;0.3621)-1\quad\text{using z tables for z=1.09}\\ &=0.7242 \end{align*}$

ans: The probability that the sample proportion will be within +-0.03 of the population proportion is 0.7242

c) The probability that the sample proportion will be within +-0.05 of the population proportion is given by the probability that $\bar{P}$ is between 0.49-0.05=0.44 and 0.49+0.05=0.54

$\begin{align*} P(0.44<\bar{P}<0.54)&=P\left(\frac{0.44-\mu_{\bar{p}}}{\sigma_{\bar{p}}} <\frac{\bar{P}-\mu_{\bar{p}}}{\sigma_{\bar{p}}}<\frac{0.54-\mu_{\bar{p}}}{\sigma_{\bar{p}}}\right )\quad\text{get the z-scores of 0.44 and 0.54}\\ &=P\left(\frac{0.44-0.49}{0.0275} <Z<\frac{0.54-0.49}{0.0275}\right )\\ &=P(-1.82<Z<1.82)\\ &=P(Z<1.82)-P(Z<-1.82)\\ &=P(Z<1.82)-P(Z>1.82)\\ &=P(Z<1.82)-(1-P(Z<1.82))\\ &=2P(Z<1.82)-1\\ &=2\times (0.5&plus;0.4656)-1\quad\text{using z tables for z=1.82}\\ &=0.9312 \end{align*}$

ans: The probability that the sample proportion will be within +-0.05 of the population proportion is 0.9312

d) When the sample size n increases, we can see that the standard error of proportion

$\sigma_{\bar{p}}=\sqrt\frac{p(1-p)}{n}=\sqrt\frac{0.49(1-0.49)}{n}$ decreases due to a larger n in the denominator.

If the standard error decrease, the z score increases, that is

z scores of 0.46 and 0.52 in part b) and 0.44 and 0.54 in part C would increase.

If the z scores increase, the probability increases.

ans: The probability would increase. This is because the increase in the sample size makes the standard error, $\sigma_{\bar{p}}$ , smaller.