Question

Suppose that 20% of the residents in a certain state support an increase in the property tax. An opinion poll will randomly sample 400 state residents and will then compute the proportion in the sample that support a property tax increase. How likely is the resulting sample proportion, ^ p , to be within 0.04 of the true proportion, p (i.e., between 0.16 and 0.24)

Answer 1

Answer:

Roughly a 95% chance.

Step-by-step explanation:

The standard deviation of sample of size 'n' and with a proportion 'p' is given by:

$SD = \sqrt{\frac{p*(1-p)}{n} }$

In this case, in a sample of 400 residents with proportion 20%, the standard deviation is:

$SD=\sqrt{\frac{0.2*(1-0.2)}{400} }\\ SD=0.02$

The upper and lower limits proposed (0.16 and 0.24) are exactly two standard deviations above and below the mean of 0.20. According to the 95% rule, this interval comprehends 95% of all data. Therefore, there is  roughly a 95% chance that the resulting sample proportion will be within 0.04 of the true proportion.

Answer 2

Answer:

P(0.16 < $\hat p$ < 0.24) = 0.9546 or 95.5%

Step-by-step explanation:

We are given that 20% of the residents in a certain state support an increase in the property tax. An opinion poll will randomly sample 400 state residents and will then compute the proportion in the sample that support a property tax increase.

Let p = % of the residents in a certain state that support an increase in the property tax = 20%

     $\hat p$ = % of the residents in a certain state that support an increase in the

            property tax in a sample of 400 residents

The one-sample z sore probability distribution for sample proportion is given by;

         Z = $\frac{\hat p -p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } }$ ~ N(0,1)   where, n = sample residents = 400

So, probability that sample proportion will lie between 0.16 and 0.24 is given by = P(0.16 < $\hat p$ < 0.24)

 P(0.16 < $\hat p$ < 0.24) = P( $\hat p$ < 0.24) - P( $\hat p \leq$ 0.16)

 P( $\hat p$ < 0.24) = P( $\frac{\hat p -p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } }$ < $\frac{0.24 -0.20}{\sqrt{\frac{0.24(1- 0.24)}{400} } }$ ) = P(Z < 1.87) = 0.96926

 P( $\hat p$ $\leq$ 0.16) = P( $\frac{\hat p -p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } }$ $\leq$ $\frac{0.16 -0.20}{\sqrt{\frac{0.16(1- 0.16)}{400} } }$ ) = P(Z < -2.18) = 1 - P(Z $\leq$ 2.18)

                                                                 = 1 - 0.98537 = 0.01463

Therefore,  P(0.16 < $\hat p$ < 0.24) = 0.96926 - 0.01463 = 0.9546

Hence, it is 95.5% likely that the sample proportion will lie between 0.16 and 0.24.