Question

A marketing manager for a cell phone company claims that more than 35% of children age 10 have cell phones. In a random sample of 5000 it is found that 1805 had cell phones. Can you conclude that the manager's claim is true? Use a significance level a=0.01. Use this scenario to answer questions 16-20. 16) State the alternative and null hypothesis. 17) Calculate the test statistic. 18) Identify the p-value. 19) Make your decision. 20) Interpret the results.

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

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

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

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

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

$z= \frac{ 0.011}{0.0067454 }$

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.

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 0.0 .5000 0.1 .5398 0.2 .5793 0.3 .6179 0.4 .6554 0.5 .6915 0.6 .7257 0.7 .7580 0.8 .7881 0.9 .8159 1.0 .8413 1.1 .8643 1.2 .8849 1.3 .9032 1.4 .9192 1.5 .9332 1.6 9152 .5040 .5438 .5832 .6217 .6591 .6950 .7291 .7611 .7910 .8186 .8438 .8665 .8869 .9049 .9207 .9345 9465 .5080 .5478 .5871 .6255 .6628 .6985 7324 7642 .7939 .8212 .8461 .8686 .8888 .9066 .9222 .9357 1.9474 5120 .5517 .5910 .6253 .6664 7019 .737 .7673 .7967 .8258 .8485 .8708 .897 .9082 .9236 .9370 .9484 .5160 .5557 .5948 .6331 .6700 .7054 .7389 .7704 .7995 .8264 .8508 .8729 .8925 .9099 .9251 .9382 .9495 .5199 .5596 .5987 .6368 .6736 .7088 .7422 .7734 .8023 .8289 .8531 .8749 .8944 .9115 .9265 9394 .9505 .5239 .5636 .6026 .6406 .6772 .7123 .7454 .7764 .8051 .8315 .8554 .8770 .8962 .9131 .9279 .9406 .9515 .5279 .5675 .6064 .6443 .6808 .7157 .7486 .7794 .8078 .8340 .8577 .8790 .8980 .9147 .9292 .9418 .9525 .5319 .5714 .6103 .6480 .6844 .7190 .7517 .7823 .8106 .8365 .8599 .8810 .8997 9162 9306 .9429 .9535 .5359 .5753 .6141 .6517 .6879 .7224 .7549 .7852 .8133 .8389 .8621 .8830 .9015 .9177 .9319 .9441 .9545

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.