Question

Refer to the information in Exercise 8.120.

a. Construct three 95% confidence intervals.

i. percent of all Asians who would welcome a white person into their families.

ii. percent of all Asians who would welcome a Latino into their families.

iii. percent of all Asians who would welcome a black person into their families.

b. Even though the three-point estimates are different, do any of the confidence intervals overlaps? Which?

c. For any intervals that do overlap, in words, what does this imply about the significance of the differences in the true proportions?

d. For any intervals that do not overlap, in words, what does this imply about the significance of the differences in the true proportions?

Short answer

(a) confidence intervals are

(i)$CI=(0.7396,0.8404)$

(ii)$CI=(0.6539,0.7661)$

(iii)$CI=(0.6014,0.7186)$

(b) Overlapping intervals$\begin{array}{r}(0.7396,0.8404)\text{and}(0.6539,0.7661)\\ (0.6539,0.7661)\text{and}(0.6014,0.7186)\end{array}$

(c) This implies that the Asian adult families can welcome a white or Latino person into their families

(d) This implies that the Asian adult families can welcome a white or Black person into their families

Step-by-step solution

part (a) (i) Explanation

i. per cent of all Asians who would welcome a white person into their families.

$p=0.79$ - sample proportion

$n=251$- sample size

Using normal approximation, we get

role="math" localid="1650867070853" $npandn(1-p)>5$

role="math" localid="1650867986307" $\Rightarrow n\ast p=198.29>5$

role="math" localid="1650867993951" $\Rightarrow n*(1-p)=52.71>5$

Taking binomial random variable as normally distributed, we get

Standard error is $\sqrt{\left(\frac{p\times (1-p)}{n}\right)}$$=0.025709$

Margin error for the confidence interval, $ME=0.0504$

$95\%$confidence interval = Sample mean $\pm$ME

role="math" localid="1650867580599" $\Rightarrow CI=0.79\pm 0.0504$

$\Rightarrow CI=(0.7396,0.8404)$

part (a) (ii) explanation

ii. per cent of all Asians who would welcome a Latino into their families.

$p=0.71$- sample proportion

$n=251$- sample size

Using normal approximation, we get

$np\text{and}n(1-p)>5\text{.}$

role="math" localid="1650868008893" $\Rightarrow n\ast p=178.21>5$

role="math" localid="1650868013813" $\Rightarrow n\times (1-p)=72.79>5$

Taking binomial random variable as normally distributed, we get

Standard error= $\sqrt{\left(\frac{0.71\times (1-0.71)}{251}\right)}$= $0.028641$

Margin error for the confidence interval ME=$0.0561$

$95\%$confidence interval = Sample mean$\pm$ME

$\Rightarrow CI=0.71\pm 0.0561$

$\Rightarrow CI=(0.6539,0.7661)$

part (a) (iii)

iii. percent of all Asians who would welcome a black person into their families.

$p=0.66$- sample proportion

$n=251$- sample size

Using normal approximation, we get

$n\times (1-p)=85.34>5$

Taking binomial random variable as normally distributed, we get

Standard error = $\sqrt{\left(\frac{0.66\times (1-0.66)}{251}\right)}$= $0.0299$

Margin error for the confidence interval ME = $0.0586$

$95\%$confidence interval = Sample mean $\pm$ME

role="math" localid="1650868650350" $\Rightarrow CI=0.66\pm 0.0586$

$\Rightarrow CI=(0.6014,0.7186)$

part (b) explanation

The overlapping confidence intervals are-$(0.7396,0.8404)\text{and}(0.6539,0.7661)$,$(0.6539,0.7661)\text{and}(0.6014,0.7186)$

part (c) explanation

We can express that there doesn't appear to be a major differentiation between the extent of Asian adults who tell that their families can invite a white individual into their families and furthermore the extent of Asian adults who say that their families can invite a Latino individual into own families.

part (d) explanation

We can express that there doesn't appear to be a major differentiation between the extent of Asian adults who tell that their families can invite a white individual into their families and furthermore the extent of Asian adults who say that their families can invite a Black individual into own families.