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 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

Answer 1

Given:

Sample size, n = 6965

Sample proportion p' =
`(1330)/(6965) = 0.19`

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:

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

`= \frac{0.19 - 0.2}{\sqrt{(0.2(1 - 0.2))/(6965)}} = -1.887`

Z = -1.887 ≈ -1.89

___________________________

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%