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Taste Test A student was tested to see if he could tell the difference between two different brands of cola. He was presented with 20 samples of cola and correctly identified 12 of the samples. Since he was correct 60 percent of the time, can we conclude that he can correctly tell the difference between two brands of cola based on this sample alone?

Short Answer

Expert verified
There is not enough statistical evidence to conclude that the student can correctly tell the difference between two brands of cola based on this sample alone.

Step by step solution

01

Null and Alternative Hypothesis Setting

The null hypothesis (H0) is that the student is guessing, meaning that the probability of choosing the correct brand (p) should be 0.5. The alternative hypothesis (H1) is that the student is not simply guessing, meaning p ≠ 0.5.
02

Statistical Test

In hypothesis testing, the normal approximation to the binomial distribution might be used with the following rules - when np and n(1 - p) are both greater than 5, the normal approximation is reasonably good. Here, n=20 and p under H0 is 0.5. Therefore np=10 and n(1-p)=10, meeting the criteria. It could be calculated that the z-score for the student's correct rate is (0.6 - 0.5) / sqrt [ 0.5 * (1 - 0.5) / 20 ] = 1.414.
03

P-value Calculation and Conclusion

A two-tail test is used here because the H1 has the form 'p ≠ 0.5'. The z-score falls into the symmetric region in the two tails of the normal curve. Look up the z-table for the z-score to find the p-value. If the p value is less than the significance level (usually 0.05), we reject H0; otherwise, we do not reject H0. In this case, the p-value related to the z-score of 1.414 exceeds 0.05, therefore, at the 0.05 significance level, we do not reject H0. This implies that there isn't enough evidence to conclude that the student can correctly tell the difference between two brands.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Null and Alternative Hypothesis
When conducting a hypothesis test, the first step is to establish the null and alternative hypotheses. The null hypothesis, denoted as \( H_0 \), represents a default position that assumes there is no effect or difference. In our scenario, this means assuming the student is guessing, so the probability of correctly identifying a cola brand is \( p = 0.5 \). This probability represents no better than random guessing. The alternative hypothesis, denoted as \( H_1 \), is what you want to test against the null. It suggests that the student's ability differs from guessing, indicating \( p eq 0.5 \). This could imply that the student can indeed distinguish between the two brands better than random chance. Hypotheses set the framework for testing whether observed results can be attributed to chance, or suggest a meaningful difference.
Normal Approximation
Normal approximation is a technique used to simplify the process of hypothesis testing for a binomial distribution. When dealing with binomial data, calculating probabilities directly can be complex. The normal approximation provides a more straightforward approach by using the normal distribution as an estimator. However, this method only works well under certain conditions.
  • The sample size \( n \) should be large enough so that both \( np \) and \( n(1-p) \) are greater than 5.

In this exercise, \( n = 20 \) and under the null hypothesis, \( p = 0.5 \). Calculating these gives \( np = 10 \) and \( n(1-p) = 10 \), both of which meet the criteria. Therefore, the normal approximation is applicable, allowing us to use a z-score to assess the hypothesis test.
Binomial Distribution
The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of trials, each with the same probability of success. This is applicable in scenarios like our cola taste test, where the student has to identify the correct brand in a series of samples. Here’s how it works:
  • Each trial is independent, meaning the outcome of one doesn't affect another.
  • The probability of success \( p \) is constant across trials.
  • The number of trials \( n \) is fixed, for example, 20 cola samples in this case.

For the test, each sample is akin to a trial, and correctly identifying a brand is a success. Understanding this distribution helps in setting our hypotheses and choosing the right statistical method for analysis.
Z-score Calculation
The z-score is a statistic used to determine how far away a sample data point is from the mean, measured in terms of standard deviation. It helps in deciding if results are statistically significant. In the context of hypothesis testing, the z-score can indicate whether the observed data would be rare if the null hypothesis were true.
To calculate the z-score in the taste test:
  • Subtract the hypothesized population proportion (\( p = 0.5 \) here) from the sample proportion (\( 0.6 \)).
  • Divide the result by the standard error of the proportion, which accounts for sample size \( n \) and variability in the data.

Mathematically, this is expressed as: \[ z = \frac{(0.6 - 0.5)}{\sqrt{ \frac{0.5 \times (1 - 0.5)}{20} }} = 1.414 \]
The calculated z-score allows us to determine the p-value and help conclude if the deviation from the null hypothesis is statistically significant.

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Most popular questions from this chapter

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