Question

In a study of cell phone usage and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 6965 subjects randomly selected from an online group involved with ears. There were 1302 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

The hypothesis test shows that we reject the null hypothesis and there is sufficient evidence to support the claim that the return rate is less than 20%

What is the claim that the return rate is less than 20% by using a statistical hypothesis method?

The claim that the return rate is less than 20% is p < 0.2. From the given information, we can compute our null hypothesis and alternative hypothesis as:

$\mathbf{H_o :p =0.2}$

$\mathbf{H_i:p < 0.2}$

Given that:
Sample size (n) = 6965

Sample proportion $\mathbf{\hat p = \dfrac{x}{n} = \dfrac{1302}{6965} \sim0.1869}$

The test statistics for this data can be computed as:

$\mathbf{z = \dfrac{\hat p - p}{\sqrt{\dfrac{p(1-p)}{n}}}}$

$\mathbf{z = \dfrac{0.1869 -0.2}{\sqrt{\dfrac{0.2(1-0.2)}{6965}}}}$

$\mathbf{z = \dfrac{-0.0131}{0.0047929}}$

z = -2.73

From the hypothesis testing, since the p < alternative hypothesis, then our test is a left-tailed test(one-tailed.

Hence, the p-value for the test statistics can be computed as:

P-value = P(Z ≤ z)

P-value = P(Z ≤ - 2.73)

By using the Excel function =NORMDIST (-2.73)

P-value = 0.00317

P-value ≅ 0.003

Therefore, we can conclude that since P-value is less than the significance level at ∝ = 0.01, we reject the null hypothesis and there is sufficient evidence to support the claim that the return rate is less than 20%

Learn more about hypothesis testing here:

brainly.com/question/15980493

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