Question

The last kiss Do people have a preference for the last thing they taste?

Researchers at the University of Michigan designed a study to find out. The researchers gave 22 students five different Hershey's Kisses (milk chocolate, dark chocolate, crème, caramel, and almond) in random order and asked the student to rate each one. Participants were not told how many Kisses they would be tasting. However, when the 5th and final Kiss was presented, participants were told that it would be their last one. $\underset{¯}{9}$Assume that the participants in the study don't have a special preference for the last thing they taste. That is, assume that the probability a person would prefer the last Kiss tasted is $p=0.20$.

a. Find the probability that 14 or more students would prefer the last Kiss tasted.

b. Of the 22 students, 14 gave the final Kiss the highest rating. Does this give convincing evidence that the participants have a preference for the last thing they taste?

Short answer

(a) The probability that 14 or more students would prefer the last Kiss tasted 0.0000

(b)There is convincing evidence that the participants have a preference for the last thing they taste.

Step-by-step solution

Part (a) Step 1: Given Information

Number of trials, $n=22$

Probability of success,$p=0.20$

Part (a) Step 2: Simplification

According to the binomial probability,

$P(X=k)=\left(\begin{array}{c}n\\ k\end{array}\right)·p^k·(1-p{)}^{n-k}$

Addition rule for mutually exclusive event:

$P(A\cup B)=P\left(A\text{or}B\right)=P\left(A\right)+P\left(B\right)$

At $k=14$,

The binomial probability to be evaluated as:

$$\begin{gathered} P(X=14)=\left(\begin{array}{c}22\\ 14\end{array}\right)·(0.20{)}^{14}·(1-0.20{)}^{22-14} \\ =\frac{22!}{14!(22-14)!}·(0.20{)}^{14}·(0.80{)}^8 \\ \approx 0.0000 \end{gathered}$$

At $k=15$,

The binomial probability to be evaluated as:

$$\begin{gathered} P(X=15)=\left(\begin{array}{c}22\\ 15\end{array}\right)·(0.20{)}^{15}·(1-0.20{)}^{22-15} \\ =\frac{22!}{15!(22-15)!}·(0.20{)}^{15}·(0.80{)}^7 \\ \approx 0.0000 \end{gathered}$$

At $k=16$,

The binomial probability to be evaluated as:

$$\begin{gathered} P(X=16)=\left(\begin{array}{c}22\\ 16\end{array}\right)·(0.20{)}^{16}·(1-0.20{)}^{22-16} \\ =\frac{22!}{16!(22-16)!}·(0.20{)}^{16}·(0.80{)}^6 \\ \approx 0.0000 \end{gathered}$$

At $k=17$,

The binomial probability to be evaluated as:

$$\begin{gathered} P(X=17)=\left(\begin{array}{c}22\\ 17\end{array}\right)·(0.20{)}^{17}·(1-0.20{)}^{22-17} \\ =\frac{22!}{17!(22-17)!}·(0.20{)}^{17}·(0.80{)}^5 \\ \approx 0.0000 \end{gathered}$$

Part (a) Step 3: Simplification

At $k=18$,

The binomial probability to be evaluated as:

$$\begin{gathered} P(X=18)=\left(\begin{array}{c}22\\ 18\end{array}\right)·(0.20{)}^{18}·(1-0.20{)}^{22-18} \\ =\frac{22!}{18!(22-18)!}·(0.20{)}^{18}·(0.80{)}^4 \\ \approx 0.0000 \end{gathered}$$

At $k=19$,

The binomial probability to be evaluated as:

$$\begin{gathered} P(X=19)=\left(\begin{array}{c}22\\ 19\end{array}\right)·(0.20{)}^{19}·(1-0.20{)}^{22-19} \\ =\frac{22!}{19!(22-19)!}·(0.20{)}^{19}·(0.80{)}^3 \\ \approx 0.0000 \end{gathered}$$

At $k=20$,

The binomial probability to be evaluated as:

$$\begin{gathered} P(X=20)=\left(\begin{array}{c}22\\ 20\end{array}\right)·(0.20{)}^{20}·(1-0.20{)}^{22-20} \\ =\frac{22!}{20!(22-20)!}·(0.20{)}^{20}·(0.80{)}^2 \\ \approx 0.0000 \end{gathered}$$

At $k=21$,

The binomial probability to be evaluated as:

$$\begin{gathered} P(X=21)=\left(\begin{array}{c}22\\ 21\end{array}\right)·(0.20{)}^{21}·(1-0.20{)}^{22-21} \\ =\frac{22!}{21!(22-21)!}·(0.20{)}^{21}·(0.80{)}^1 \\ \approx 0.0000 \end{gathered}$$

At $k=22$,

The binomial probability to be evaluated as:

$$\begin{gathered} P(X=22)=\left(\begin{array}{l}22\\ 22\end{array}\right)·(0.20{)}^{22}·(1-0.20{)}^{22-22} \\ =\frac{22!}{14!(22-14)!}·(0.20{)}^{22}·(0.80{)}^0 \\ \approx 0.0000 \end{gathered}$$

Since two different numbers of successes are impossible on same simulation.

Apply addition rule for mutually exclusive events:

$$\begin{gathered} P(X\le 14)=P(X=14)+P(X=15)+P(X=16)+P(X=17) \\ +P(X=18) \\ +P(X=19)+P(X=20)+P(X=21) \\ +P(X=22)\backslash \backslash =0.0000+0.0000+0.0000+0.0000+0.0000 \\ +0.0000\backslash \backslash +0.0000+0.0000+0.0000 \\ =0.0000 \\ =0.00\% \end{gathered}$$

Part (b) Step 1:Given Information

Number of trials, $n=22$

Probability of success,$p=0.20$

Part (b) Step 2:Simplification

When the probability is less than $0.05$, it is considered to be small.

In this scenario, keep in mind that the likelihood is low enough.

As a result, it seems doubtful that 14 of the 22 students rated the final kiss the highest rating.

This means that there is compelling evidence that the participants prefer the last thing they taste.