Answer: z= 3.55
Step-by-step explanation:
Let p be the population proportion of the asthmatic children requires emergency room care for asthmatic complications at some time in the 6 month period.
As per give , we have
[tex]H_0:p=0.152\\\\ H_a: p\neq 0.152[/tex]
Since alternative hypothesis is two-tailed , so test is a two-tailed test.
To test their claim, researcher took a SRS of 260 asthmatic children and had them use a steroidal inhaler for 6 months. Of the 260 children, 60 required emergency room care for asthmatic complications at some time in the 6 month period.
Sample size : n= 260
Sample proportion : [tex]\hat{p}=\dfrac{60}{260}\approx0.231[/tex]
Test statistic : [tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]
[tex]z=\dfrac{0.231-0.152}{\sqrt{\dfrac{0.152(1-0.152)}{260}}}\approx 3.54808524246\approx3.55[/tex]
∴ The test statistic : z= 3.55