Question

A survey was sent out to compare the proportion of adults who use their car horns when driving for two age populations (1=younger adults, defined as between 20 and 39 years old and 2 =older adults, defined as over 60 years old). The following data was obtained from those who responded.
Calculate the 90% confidence interval using the standard normal distribution. Note that 1 =0.52. P2= 0.35, and s.e.(P1-P2) =0.0338. Round to the fourth decimal point. Please enter you answer in the following format: (lower value, upper value)

Use the horn Use the horn
Group Yes No Total
1= younger adults 261 240 501
2= older adults 123 229 352

Answer 1

Answer:

The  90% confidence interval for the difference between proportions is (0.115, 0.228).

As the value 0 is not included in the interval, we can conclude that there is significant difference in the proportion of youger adults that use the horn and older adults that use the horn.

Step-by-step explanation:

We want to calculate the bounds of a 90% confidence interval.

For a 90% CI, the critical value for z is z=1.645.

The sample 1 (younger adults) , of size n1=501 has a proportion of p1=0.521.

$p_1=X_1/n_1=261/501=0.5210$

The sample 2 (older  adults), of size n2=352 has a proportion of p2=0.3494.

$p_2=X_2/n_2=123/352=0.3494$

The difference between proportions is (p1-p2)=0.1715.

$p_d=p_1-p_2=0.5210-0.3494=0.1715$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{261+123}{501+352}=\dfrac{384}{853}=0.4502$

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.4502*0.5498}{501}+\dfrac{0.4502*0.5498}{352}}\\\\\\s_{p1-p2}=\sqrt{0.0005+0.0007}=\sqrt{0.0012}=0.0346$

Then, the margin of error is:

$MOE=z \cdot s_{p1-p2}=1.645\cdot 0.0346=0.0569$

Then, the lower and upper bounds of the confidence interval are:

$LL=(p_1-p_2)-z\cdot s_{p1-p2} = 0.1715-0.0569=0.115\\\\UL=(p_1-p_2)+z\cdot s_{p1-p2}= 0.1715+0.0569=0.228$

The  90% confidence interval for the difference between proportions is (0.115, 0.228).