Answer: We will reject the null hypothesis and it is true that more than 50% of U.S tax returns were filled electronically last year.
Step-by-step explanation:
Since we have given that
[tex]H_0:p=\hat{p}\\\\H_a:\hat{p}>p[/tex]
Here, p = 0.5
n = 150
x = 86
so, [tex]\hat{p}=\dfrac{x}{n}=\dfrac{86}{150}=0.573[/tex]
So, test statistic value would be
[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\z=\dfrac{0.573-0.50}{\sqrt{\dfrac{0.5\times 0.5}{150}}}\\\\z=1.7963[/tex]
At 5% level of significance, z = 1.645
So, 1.645<1.7963
Hence, we will reject the null hypothesis.
P-value would be
[tex]P(z>1.7963)\\\\=1-P(z\leq 1.7963)\\\\=1-0.9638\\\\=0.0362[/tex]
Hence, we will reject the null hypothesis and it is true that more than 50% of U.S tax returns were filled electronically last year.