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 1330 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. A. Upper H 0: pequals0.2 Upper H 1: pless than0.2 B. Upper H 0: pgreater than0.2 Upper H 1: pequals0.2 C. Upper H 0: pequals0.2 Upper H 1: pnot equals0.2 D. Upper H 0: pnot equals0.2 Upper H 1: pequals0.2 E. Upper H 0: pless than0.2 Upper H 1: pequals0.2 F. Upper H 0: pequals0.2 Upper H 1: pgreater than0.2 The test statistic is zequals nothing. (Round to two decimal places as needed.) The P-value is nothing. (Round to three decimal places as needed.) Because the P-value is

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Chrisnando Chrisnando · Answer 1 · 2020-06-04T02:51:50+02:00

Answer:

Given:

Sample size, n = 6965

Sample proportion p' = [tex] \frac{1330}{6965} = 0.19 [/tex]

Let's claim that the return rate is less than 20%, ie, P = 0.2

Significance level = 0.01

The null and alternative hypotheses:

H0 : P = 0.2

H1 : P < 0.2

Option A is correct

_____________________________

For the test statistic, Z, let's use the formula:

[tex] Z = \frac{p'- P}{\sqrt{\frac{P(1-P)}{n}}} [/tex]

[tex] = \frac{0.19 - 0.2}{\sqrt{\frac{0.2(1 - 0.2)}{6965}}} = -1.887 [/tex]

Z = -1.887 ≈ -1.89

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The p-value.

This is a left tailed test.

Using z table,

P-value = P(Z ≤ -1.89) = 0.029

The pvalue is 0.029

Decision:

Because the pvalue, 0.029 is greater than level of significance, 0.01, we fail to reject null hypothesis, H0.

Conclusion:

There is not enough sufficient evidence to conclude that the return rate is less than 20%