 Maria Huey

2021-12-16

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 confirm the belief that one design is just as likely to be selected as another? Explain.
$\begin{array}{|cc|}\hline \text{Design}& \text{Number of Times preferred}\\ 1& 5\\ 2& 15\\ 3& 30\\ 4& 40\\ 5& 10\\ \hline\end{array}$ Andrew Reyes

Step 1
The probability is the number of favorable outcomes divided by the number of possible outcomes:
$P\text{(outcomes)}=\frac{\mathrm{#}\text{of favorable outcomes}}{\mathrm{#}\text{of possible outcomes}}=\frac{1}{5}=0.2=20\mathrm{%}$
It does not appear that the data confirm the belief that one desing is just as likely to be selected as another, because design 4 is preferred 40 times, while design 1 is only preferred 5 times and this is a very large difference in a group of 100. accimaroyalde

Step 1
The assumption made is that all the 5 different packages are equally likely, i.e. the probability of selecting a package is $\frac{1}{5}=0.20$
The probability distribution is shown below.
According to the probability distribution:
The probability of a person preferring design 1 is,
$P\left(X=1\right)=0.10$
The probability of a person preferring design 2 is,
$P\left(X=2\right)=0.05$
The probability of a person preferring design 3 is,
$P\left(X=3\right)=0.30$
The probability of a person preferring design 4 is,
$P\left(X=4\right)=0.40$
The probability of a person preferring design 1 is,
$P\left(X=5\right)=.15$
So it can be seen that the probability of preferring any of the 5 designs are not same.
Thus, the designs are not equally likely.
The correct option is "No, a probability of about 0.20 would be assigned using the relative frequency method if selection is equally likely."
$\begin{array}{|ccc|}\hline \text{Design}\left(X\right)& \text{number of times Preferred}\left(f\left(x\right)\right)& P\left(X\right)\\ 1& 10& 0.10\\ 2& 5& 0.05\\ 3& 30& 0.30\\ 4& 40& 0.40\\ 5& 15& 0.15\\ Total=& 100& 1.00\\ \hline\end{array}$ nick1337

Step 1
Given:
Probability (one number) $=\frac{1}{5}=0.2$
$\begin{array}{|cc|}\hline \text{Design}& \text{Time preferred}\\ 1& 5\\ 2& 15\\ 3& 30\\ 4& 40\\ 5& 10\\ & \sum =100\\ \hline\end{array}$
Step 2
$\text{Expected value}=np$
$=100×0.2$
$=20$
Null hypothesis : ${H}_{0}:$ all designs preference is same
Alternative hypothesis : ${H}_{a}:$ all designs preference is not same
It is a two tailed test
Step 3
Apply chi square test:
${X}^{2}=\frac{\left(O-E{\right)}^{2}}{E}$
$=\frac{\left(5-20{\right)}^{2}}{20}+\frac{\left(15-20{\right)}^{2}}{20}+\frac{\left(30-20{\right)}^{2}}{20}+\frac{\left(40-20{\right)}^{2}}{20}+\frac{\left(10-20{\right)}^{2}}{20}$
$=11.25+1.25+5+20+5$
Step 4
Degree of freedom $=5-1=4$
$\text{significance level}=0.05$
$\text{p-value}=0$
$\text{As p-value}<\text{significance level}$
so, it rejects null hypothesis
We have insufficient evidence to conclude that one design is just as likely to be selected as another.

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