- – In existing experimentation, $100$ customers were asked to choose the design they liked. The subsequent data were acquired. Do the data demonstrate the thought that one design is just as conceivable to be designated as another? Explain.
Figure-1
This problem aims to familiarize us with the concept of the null hypothesis and probability distribution. The concept of inferential statistics is used to explain the problem, in which the null hypothesis helps us test different relationships among different phenomena.
In mathematics, the null hypothesis, directed to as $H_0$, declares that the two occurring prospects are exact. Whereas the probability distribution is a statistical procedure that represents all the potential values and possibilities that a spontaneous variable can handle within a provided range.
Expert Answer
According to the given statement, the null hypothesis $H_0$ can be obtained as; all the designs are just as likely to be selected as any other design, whereas the alternative hypothesis $H_a$ can be counter positive of the above statement, that is all designs are not given the same preference, then the probability of selecting a single package can be given as:
\[ P(X) = \dfrac{1}{5} = 0.20 \]
But according to the probability distribution, we can achieve the following results:
The probability that the first design gets chosen is,
\[ P(X = 1) = 0.05 \]
The probability that the second design gets chosen is,
\[ P(X = 2) = 0.15 \]
The probability that the third design gets chosen is,
\[ P(X = 3) = 0.30 \]
The probability that the fourth design gets chosen is,
\[ P(X = 4) = 0.40 \]
The probability that the fifth design gets chosen is,
\[ P(X = 3) = 0.10 \]
Figure-2
Hence, from the above probability distribution, we can notice that the probability of choosing any of the above $5$ designs is not the same.
Thus the designs are not just as equally likely to each other hence rejecting our null hypothesis. In order to make the selection to be equally likely, a probability of about $0.20$ would be assigned using the relative frequency distribution method.
Numerical Result
The probability of choosing any of the given $5$ designs is not the same. Thus, the designs are not just as equally likely to each other, hence it rejects the null hypothesis.
Example
Consider that a sample space has $5$ equally likely practical results, $E_1, E_2, E_3, E_4, E_5$, let,
\[ A = [E_1, E_2] \]
\[B = [E_3, E_4] \]
\[C = [E_2, E_3, E_5] \]
Find the probability of $A$, $B$, $C$, and $P(AUB)$.
Following are the probabilities of $A$, $B$, and $C$:
\[ P(A) = P(E_1, E_2) = \dfrac{2}{5} = 0.4 \]
\[ P(B) = P(E_3, E_4) = \dfrac{2}{5} = 0.4 \]
\[ P(C) = P(E_2, E_3, E_5) = \dfrac{3}{5} = 0.6 \]
Probability of $AUB$:
\[ P(AUB) = P(A) + P(B) \]
\[ P(AUB) = P(E_1, E_2) + P(E_3, E_4)\]
\[P(AUB) = P(E_1, E_2, E_3, E_4)\]
\[P(AUB) = \dfrac{4}{5} \]
\[P(AUB) = 0.80 \]