91影视

It is given that $x_1=450$

$x_2=351$

$n_1=3000$

$n_2=1800$

$伪=0.01$

Claim: Population of all teens with hearing loss is higher in$2010$

The hypothesis is:

Null: $H_0:p_1=p_2$

Alternative: $H_a:p_1<p_2$

$p_1$is proportion of teens in $1990$with hearing loss and $p_2$is \text { proportion of teens in $2010$with hearing loss.

Conditions are:

Random: Teenagers are independent random samples.

Independent: $3000<10\%\mathrm{in}1990$and $1800<10\%\mathrm{in}2010$

Normal: Success are $450,351$and failures are $3000-450=2550$and $1800-351=1449$. All are greater than ten.

All condition are satisfied. Hypothesis test can be performed.

Sample Proportion: ${\hat{p}}_1=\frac{x_1}{n_1}=\frac{450}{3000}=0.15$

${\hat{p}}_2=\frac{x_2}{n_2}=\frac{351}{1800}=0.195$

and ${\hat{p}}_p=\frac{x_1+x_2}{n_1+n_2}=\frac{450+351}{3000+1800}=\frac{800}{4800}=0.1667$

Test statistics:

$z=\frac{{\hat{p}}_1-{\hat{p}}_2-\left(p_1-p_2\right)}{\sqrt{{\hat{p}}_p\left(1-{\hat{p}}_p\right)}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{0.15-0.195-0}{\sqrt{0.1667(1-0.1667)}\sqrt{\frac{1}{3000}+\frac{1}{1800}}}鈮�-4.05$

Probability: $P=P(Z<-4.05)鈮�0$

Also, $P<0.01鈬�\text{Reject}{H}_0$

So, there is convincing evidence that the proportion of all teens with hearing loss has increased over the years.

As $P鈮�0$

There is a very small chance of obtaining similar test results or more extreme when the proportion of all teens with hearing loss is greater in $2010$ compared to $1990$ .