Question

Determine whether a normal sampling distribution can be used for the following sample statisics If it oan be used, test the claim about the difference between two population proportions p, and P2 at the level of significance a Assume that the samples are random and independent The samples are random and Independent A normal sampling distribultion (Round to two decimal places as needed.) can be used because n,p- 28.93,n-41.0 ngp 3306 nd 24-46.94 Sale กเติ and alternativo hypot eses, ส applicable. D The conditions to use a normal sampling distribution are not met Calculate the standardized test statistic for the difference p,-Pz if applioable. Select the correct choloe below and i ncessary,f i the anawer box to complete your choice Round to two decimal places as needed.) B The conditions to use a normal sampling distribution are not met

Answer 1

The sample proportions are ,

$p_1=\frac{x_1}{n_1}=\frac{32}{70}=0.4571$

$p_2=\frac{x_2}{n_2}=\frac{30}{80}=0.3750$

The pooled estimate of the proportion is ,

$\bar{p}=\frac{x_1+x_2}{n_1+n_2}=\frac{30+32}{70+80}=0.4133$

$\bar{q}=1-\bar{p}=0.5867$

Since , the samples are random and independent.

A normal sampling distribution can be used because $n_1\bar{p}=70*0.4133=28.93$ , $n_1\bar{q}=70*0.5867=41.07$ , $n_2\bar{p}=80*0.4133=33.06$ and $n_2\bar{q}=80*0.5867=46.94$

Hypothesis : $H_0:p_1=p_2$ Vs $H_a:p_1\neq p_2$

The standardized test statistic is ,

$Z=\frac{p_1-p_2}{\sqrt{\bar{p}\bar{q}(\frac{1}{n_1}+\frac{1}{n_2})}}\to N(0,1)$

$=\frac{0.4571-0.3750}{\sqrt{0.4133*0.5867(\frac{1}{70}+\frac{1}{80})}}=1.02$