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A marketing manager for a cell phone company claims that more than 35% of children age 10 have cell phones. In a random sampl
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Answer #1

Solution:
Given:

p = 0.35

n = 5000

x = 1805

Claim: More than 35% of children age 10 have cell phones. that is: p > 0.35

Level of significance = alpha = 0.01

Question 16)
Alternative hypothesis:

Since claim is directional to right ( > ) side, this is right tailed test.

Thus

H1: p > 0.35

Null hypothesis:

H0: p = 0.35

Question 17) Test statistic:

z= \frac{\hat{p}-p}{\sqrt{\frac{p\times (1-p)}{n}}}

where

\hat{p}=\frac{x}{n} = \frac{1805}{5000}=0.361

thus

z= \frac{\hat{p}-p}{\sqrt{\frac{p\times (1-p)}{n}}}

z= \frac{0.361 -0.35}{\sqrt{\frac{0.35\times (1-0.35)}{5000}}}

z= \frac{ 0.011}{\sqrt{\frac{0.35\times 0.65 }{5000}}}

z= \frac{ 0.011}{\sqrt{ 0.0000455 }}

z= \frac{ 0.011}{0.0067454 }

z= 1.63

Question 18) P-value:

For right tailed test , P-value is:

P-value = P(Z > z test statistic)

P-value = P(Z > 1.63)

P-value = 1 - P(Z < 1.63)

Look in z table for z = 1.6 and 0.03 and find corresponding area.

P( Z< 1.63) = 0.9484

thus

P-value = 1 - P(Z < 1.63)

P-value = 1 - 0.9484

P-value = 0.0516

Question 19) Decision:

Decision Rule:
Reject null hypothesis H0, if P-value < 0.01 level of significance, otherwise we fail to reject H0

Since P-value = 0.0516 > 0.01 level of significance, we fail to reject H0.

Question 20) Interpret the results:

At 0.01 level of significance, we do not have sufficient evidence to support the marketing manager's claim that more than 35% of children age 10 have cell phones.

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