Question

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.

Answer 1

Answer:

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%.