A poll found that in a random sample of 1000 American adults, 710 indicated that they oppose the reinstatement of a military draft. Is there convincing evidence that the proportion of American adults who oppose reinstatement of the draft is greater than 0.7? Use a significance level of 0.05. (Round your test statistic to two decimal places and your P-value to four decimal places.)

Respuesta :

Answer:

-Accept [tex]H_o[/tex]

-There is no sufficient evidence to back the claim that the proportion of American adults who oppose reinstatement of the draft is greater than 0.7

Step-by-step explanation:

-We state our hypotheses as:

[tex]H_o:p=0.7\\\\\\H_a:p>0.7[/tex]

#We the calculate the proportion to be studied using the formula:

[tex]\hat p=\frac{x}{n}\\\\=\frac{710}{1000}\\\\=0.71[/tex]

-The z-value from this proportion is calculated as follows:

[tex]z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}\\\\=\frac{0.71-0.7}{\sqrt{\frac{0.7\times0.3}{1000}}}\\\\\\=0.6901[/tex]

#We compare this z-value to the z-value at 95% confidence level:

[tex]z_{0.025}=1.645\\\\0.6901<1.645[/tex]

Hence, Accept the null hypothesis.

-There is no sufficient evidence to back the claim that the proportion of American adults who oppose reinstatement of the draft is greater than 0.7