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In the book Essentials of Marketing Research, William R. Dillon, Thomas J. Madden, and Neil H. Firtle discuss a research proposa

MakcuM [25]

Answer:

Null hypothesis: $p_{1} = p_{2}$

Alternative hypothesis: $p_{1} \neq p_{2}$

$z=\frac{0.179-0.15}{\sqrt{0.17(1-0.17)(\frac{1}{140}+\frac{1}{60})}}=0.500$

$p_v =2*P(Z>0.500)=0.617$

So the p value is a very low value and using any significance level for example $\alpha=0.05, 0,1,0.15$ always $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the two proportions NOT differs significantly.

Step-by-step explanation:

Data given and notation

$X_{1}=25$ represent the number of homeowners who would buy the security system

$X_{2}=9$ represent the number of renters who would buy the security system

$n_{1}=140$ sample 1

$n_{2}=60$ sample 2

$p_{1}=\frac{25}{140}=0.179$ represent the proportion of homeowners who would buy the security system

$p_{2}=\frac{9}{60}= 0.15$ represent the proportion of renters who would buy the security system

z would represent the statistic (variable of interest)

$p_v$ represent the value for the test (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the two proportions differs , the system of hypothesis would be:

Null hypothesis: $p_{1} = p_{2}$

Alternative hypothesis: $p_{1} \neq p_{2}$

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$ (1)

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{25+9}{140+60}=0.17$

Calculate the statistic

Replacing in formula (1) the values obtained we got this:

$z=\frac{0.179-0.15}{\sqrt{0.17(1-0.17)(\frac{1}{140}+\frac{1}{60})}}=0.500$

Statistical decision

For this case we don't have a significance level provided $\alpha$ , but we can calculate the p value for this test.

Since is a two sided test the p value would be:

$p_v =2*P(Z>0.500)=0.617$

So the p value is a very low value and using any significance level for example $\alpha=0.05, 0,1,0.15$ always $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the two proportions NOT differs significantly.

6 0

3 years ago

What is the length of the hypotenuse in the traingle shown above?

il63 [147K]

Answer:

17 yards

Step-by-step explanation:

X = √[(15^2) + (8^2)]

x = √[225 + 64]

x = √289

x = 17

8 0

2 years ago

Craig has a 6 ft by 6 ft square garden in his backyard. He wants to transform the garden into a rectangular shape and reduce the

Yanka [14]

X = 2
So 6 + 2 = 8
6 - 2 = 4
8 x 4 = 32

(6-x)(6+x) = 32
36 - x^2 = 32
-36. -36
-x^2 = -4
X = 2

7 0

3 years ago

According to the University of Nevada Center for Logistics Management, 6% of all merchandise sold in the United States gets retu

jeka57 [31]

Answer:

$p_v =P(z>3.390)=0.000349$

If we compare the p value and the significance level given $\alpha=0.1$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the proportion of returns at the Houston store is significantly higher than 0.06 or 6%.

Step-by-step explanation:

1) Data given and notation n

n=80 represent the random sample taken

X=12 represent number of items returned

$\hat p=\frac{12}{80}=0.15$ estimated proportion of items returned

$p_o=0.06$ is the value that we want to test

$\alpha=0.1$ represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that proportion of returns at the Houston store was more than the national expectation (0.06):

Null hypothesis: $p\leq 0.06$

Alternative hypothesis: $p >0.06$

When we conduct a proportion test we need to use the z statistic, and the is given by:

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

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.15 -0.06}{\sqrt{\frac{0.06(1-0.06)}{80}}}=3.390$

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.1$ . The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(z>3.390)=0.000349$

If we compare the p value and the significance level given $\alpha=0.1$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the proportion of returns at the Houston store is significantly higher than 0.06 or 6%.

4 0

3 years ago

Berniece earns $134 each day she works. Every day she spends $8 on breakfast and $12 on lunch. How much money will she have in 2

Svetradugi [14.3K]

25 days = 2850$
50 days =5700$

7 0

2 years ago

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