Answer
$-z_{\frac{α}{2}}\lt z_0\lt z_{\frac{α}{2}}$: null hypothesis is not rejected.
There is not enough evidence to conclude that the data contradicts the results of Toluna.
Work Step by Step
$H_0:~p=0.26$ versus $H_1:~p\ne0.26$
Requirement:
$np_0(1−p_0)=60\times0.26(1-0.26)=11.544\gt10$
$p̂ =\frac{x}{n}=\frac{21}{60}=0.35$
$z_0=\frac{p̂ -p_0}{\sqrt {\frac{p_0(1-p_0)}{n}}}=\frac{0.35-0.26}{\sqrt {\frac{0.26(1-0.26)}{60}}}=1.59$
Two-tailed test.
Using the classical method:
$z_{\frac{α}{2}}=z_{0.025}$
If the area of the standard normal curve to the right of $z_{0.025}$ is 0.025, then the area of the standard normal curve to the left of $z_{0.025}$ is $1−0.025=0.975$
According to Table V, the z-score which gives the closest value to 0.975 is 1.96.
Also, $-z_{\frac{α}{2}}=-z_{0.025}=-1.96$
Since $-z_{\frac{α}{2}}\lt z_0\lt z_{\frac{α}{2}}$, we do not reject the null hypothesis.
