In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6970 subjects randomly selected from an online group involved with ears. There were 1334 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.

we will fail to reject the null hypothesis and conclude that the return rate is less than 20%.

Step-by-step explanation:

We are given;

Sample size;n = 6970

Success rate;X = 1334/6970 = 0.1914

Now, we want to test the claim that the return rate is less than p = 0.2, hence the null and alternative hypothesis are respectively;

H0: μ < 0.2

Ha: μ ≥ 0.2

The standard deviation formula is;

σ = √(x(1 - x)/n)

σ = √(0.1914(1 - 0.1914)/6970)

σ = 0.004712

Now for the test statistic, formula is;

z = (x - μ)/σ

z = (0.1914 - 0.2)/0.004712

z = -1.825

From the a-distribution table attached, we have a value of 0.03362.

This p-value gotten from the z-table is more than the significance value of 0.01. Thus, we will fail to reject the null hypothesis and conclude that the return rate is less than 20%.

2

step-by-step explanation:

m∠5 + m∠3 = m∠1   is true.

m∠5 + m∠6 = 180 is true

m∠2 +m∠3 = m∠6   is true

m∠2 +m∠3 +m∠5 = 180   is true