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Tanya [424]
3 years ago
10

A company that manufactures toothpaste is studying five different package designs. Assuming that one design is just as likely to

be selected by a consumer as any other design, what selection probability would you assign to each of the package designs? In an actual experiment, 100 consumers were asked to pick the design they preferred. The following data were obtained. Do the data appear to confirm the belief that one design is just as likely to be selected as another? Explain design number of times preferred.
1 5
2 15
3 30
4 40
5 10
Mathematics
1 answer:
Pachacha [2.7K]3 years ago
4 0

Answer:

The selection probability that would be assigned to each of the package designs , since they are equally likely is 1/5= 0.2

step-by-step explanation:

the data appear not to confirm the belief that one design is just as likely to be selected as another because 1 is preferred 5 times and 4 is preferred 40 times, this is very wide apart, and it confirmed the belief that one design is not as likely to be selected as another.

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Answer:

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Step-by-step explanation:

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Answer:

Case n =5

z=\frac{4.8-5}{\frac{0.5}{\sqrt{5}}}=-0.894  

Case n =15

z=\frac{4.8-5}{\frac{0.5}{\sqrt{15}}}=-1.549  

Case n = 40

z=\frac{4.8-5}{\frac{0.5}{\sqrt{40}}}=-2.530  

P value

Case n =5

p_v =P(z  

Case n =15

p_v =P(z  

Case n =40

p_v =P(z  

Step-by-step explanation:

Data given and notation  

\bar X=4.8 represent the sample mean

\sigma=0.5 represent the population standard deviation

n sample size  

\mu_o =5 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

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

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is lower than 5, the system of hypothesis would be:  

Null hypothesis:\mu \geq 5  

Alternative hypothesis:\mu < 5  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

Case n =5

z=\frac{4.8-5}{\frac{0.5}{\sqrt{5}}}=-0.894  

Case n =15

z=\frac{4.8-5}{\frac{0.5}{\sqrt{15}}}=-1.549  

Case n = 40

z=\frac{4.8-5}{\frac{0.5}{\sqrt{40}}}=-2.530  

P-value  

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

Case n =5

p_v =P(z  

Case n =15

p_v =P(z  

Case n =40

p_v =P(z  

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Answer:

Step-by-step explanation:

Its like doing a long division with base 10 numbers:

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