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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.
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: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 = 6970Success rate;X = 1334/6970 = 0.1914Now, 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 < 0.2Ha: μ ≥ 0.2The standard deviation formula is;σ = √(x(1 - x)/n)σ = √(0.1914(1 - 0.1914)/6970)σ = 0.004712Now for the test statistic, formula is;z = (x - μ)/σz = (0.1914 - 0.2)/0.004712z = -1.825From 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%....