Question

886 randomly sampled teens were asked which of several personal items of information they thought it okay to share with someone they had just met. 44% said it was okay to share their e-mail addresses, but only 29% said they would give out their cell phone numbers. A researcher claims that a 2-proportion z-test could tell whether there was a real difference among all teens. Explain why that test would not be appropriate for these data.

Answer 1

Answer:

 Test statistic $Z = 6.5217$

The Calculated value Z =6.5217 > 1.96 at 5% level of significance

Null hypothesis is rejected

There is real difference among all teens

Step-by-step explanation:

Step(i):-

Given large sample size 'n' = 886

first sample proportion p⁻₁ = 44% =0.44

Second sample proportion p⁻₂ = 29% =0.29

Null hypothesis:H₀:There is no significant difference between two proportions

Alternative Hypothesis:H₁: There is significant difference between two proportions

Level of significance ∝ = 0.95 or 95%

$Z_{\frac{\alpha }{2} = Z_{\frac{0.05}{2} } = Z_{0.025} } =1.96$

Step(ii):-

Test statistic

$Z = \frac{p^{-} _{1}-p^{-} _{2} }{\sqrt{pq(\frac{1}{n_{1} } +}\frac{1}{n_{2} } )}$

Where    'p'

$p = \frac{n_{1} p^{-} _{1}+n_{2}p^{-} _{2} }{n_{1} +n_{2} }$

$p = \frac{886 (0.44)+886(0.29) }{886+886 }$

p =   0.365

q = 1-p =1-0.365 =0.635

$Z = \frac{0.44-0.29 }{\sqrt{0.365(0.635)(\frac{1}{886} +}\frac{1}{886} )}$

$Z = \frac{0.15}{ 0.023} = 6.5217$

The Calculated value Z =6.5217 > 1.96 at 5% level of significance

Conclusion:-

Null hypothesis is rejected

There is real difference among all teens