Instruction for Hypothesis Testing (Problems 17-25, Except 24) For One Population: Show your work for finding...

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Instruction for Hypothesis Testing (Problems 17-25, Except 24) For One Population: Show your work for finding Test value, Cri 19. (8 points) A telephone company wants to advertise that more than 30% of all its customers have more than two telephones.

math Statistics-And-Probability

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Answer 1

Answer #1

(a) Parameters of test

A parameter is a characteristic of a population, and in given question we have given information about population is company believes than more 30 % of its customers uses more than two telephones, So

population proportion = p = 0.30

(b)Claim of the company is that more than 30% of its customers have more than two telephones

(c)Given information

Sample Size = n = 200

Number of people using more than two phones = X = 72

Sample proportion

$p' = \frac{X}{n} = \frac{72}{200} = 0.36$

Sample proportion = p' = 0.36

Level of significance = α = 0.05

(1) Null and alternative hypothesis need to be tested

Ho: p ≤ 0.30

Ha: p > 0.30 (claim)

This corresponds to a right-tailed test

(2) Critical value

zc is calculated using excel formula =NORM.S.INV(1-0.05)

zc = 1.96

I have also shown critical value determination by table method

Number in the table represents PIZsz) 02 Z 0.05 0.02 .5080 .5478 0.0 0.1 0.08 .5319 .5199 .5596 .5714 0.01 .5040 .5438 .5832 Test value for p value method

$z = \frac{p'-p}{\sqrt{\frac{p(1-p)}{n}}} = \frac{0.36-0.30}{\sqrt{\frac{0.30(1-0.30)}{200}}}= 1.852$

So z test = 1.852

(3)p value corresponding to z test is calculated as p = 0.0322 using excel formula

=1-NORM.S.DIST(1.852,TRUE)

using table we find p value for z= 1.85 and subtract from 1 for finding p value for right tailed test

From table we have p value = 1-0.9678 = 0.0322

(4) bell curve showing critical and test value Where critical region is above z = 1.64 and white region is non critical region

Comparing p value with level of significance

p = 0.0322 which is less that α=0.05 So null hypothesis is rejected

it can also be observed that z = 1.85 > zc = 1.64, it is then concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to support the claim that the customers of company are using more than two telephones at the α=0.05 significance level.

Note - Kindly go through the Solution , I have tried best to put all the answers serially.

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