In a study of cell phone usage and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 6967 subjects randomly selected from an online group involved with ears. There were 1308 surveys returned. Use a 0.01 significance level to test the claim that the return rate is less than​ 20%. Use the​ P-value method and use the normal distribution as an approximation to the binomial distribution. Identify the null hypothesis and alternative hypothesis.

Respuesta :

Answer:

Following are the response to the given question:

Step-by-step explanation:

Testing hypothesis:

[tex]H_0: p \geq 0.2 \\\\H_a: p < 0.2[/tex]

When it's a lower tailed test:

[tex]\hat{p}=\frac{x}{n}=\frac{1308}{6967}\approx 0.1877\\\\n= 6967\\\\[/tex]

claimed proportion, [tex]P = 0.2[/tex]

Significance level, [tex]\alpha= 0.01[/tex]

calculating statistics :  

[tex]\hat{p}, \sigma_{\hat{p}}=\sqrt{\frac{P \times (1-p)}{n}}\\\\[/tex]

        [tex]=\sqrt{\frac{0.2 \times (1-0.2)}{6967}}\\\\ =\sqrt{\frac{0.2 -0.04}{6967}}\\\\ =\sqrt{\frac{0.16}{6967}} \\\\ =\sqrt{2.2965}\approx 1.51[/tex]