Suppose You Were Hired To Determine If College Students Coul

Suppose You Were Hired To Determine If College Students Could Tell The

Suppose you were hired to determine if college students could tell the difference in taste between Sprite and Sierra Mist. You conducted a study by randomly sampling 90 students from a large lecture on the Oregon State University campus. Each student was individually brought to a tasting room where three cups with lids were placed in front of them. Two of the cups contained one of the brands of soda (say Sprite) and the other cup contained the other brand of soda (Sierra Mist). Students did not know which brand was in which cup.

Using a straw that was placed through the lid of each cup, each student was to taste the soda in each cup and report to you which cup they thought contained the different brand. Suppose 52 of the 90 students were able to correctly identify the cup that contained the different brand of soda. Is that evidence that college students could tell the difference in taste between Sprite and Sierra Mist? To answer this question of interest, answer the following questions.

Paper For Above instruction

The primary aim of this study is to determine whether college students can distinguish between the tastes of Sprite and Sierra Mist. To address this, a sample of 90 students from Oregon State University participated in a blind taste test, and the outcomes provide data to analyze whether a significant difference exists from what would be expected under random guessing.

Identification of the random variable and its distribution

The random variable in this study is the number of students who correctly identify the different brand of soda out of the total sample size. This variable follows a binomial distribution because each student's response is a Bernoulli trial—either they correctly identify the different soda or they do not, with fixed probability p, and the total number of students is fixed at 90. Each trial is assumed independent, making the binomial model appropriate.

Interpretation of p in context

The parameter p represents the true proportion of all college students who can correctly distinguish between the taste of Sprite and Sierra Mist in the population from which the sample was drawn. If p is significantly greater than 0.5, it suggests that students can reliably identify the different soda brands beyond what would be expected by chance.

Sample proportion calculation

The sample proportion \(\hat{p}\) is calculated as:

\(\hat{p} = \frac{\text{Number of students who identified correctly}}{\text{Total number of students}}\)

Substituting the values:

\(\hat{p} = \frac{52}{90} ≈ 0.5778\)

Hypotheses formulation

  1. Null hypothesis \(H_0: p = 0.5\)
  2. Alternative hypothesis \(H_A: p > 0.5\)

In words: The null hypothesis states that students are guessing randomly—no ability to distinguish tastes—while the alternative posits that students can correctly identify the different soda at a rate greater than chance.

Choosing the appropriate hypothesis test

Assuming independence of observations and a sample size sufficiently large for approximation, a one-proportion z-test is appropriate for testing the proportion against the null value of 0.5. This test assesses whether the observed sample proportion significantly exceeds 0.5.

Performing the hypothesis test

The test statistic is calculated as:

\(z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}}\)

where \(p_0 = 0.5\), \(\hat{p} ≈ 0.5778\), and \(n=90\).

Calculating the standard error:

\(\text{SE} = \sqrt{\frac{0.5 \times 0.5}{90}} ≈ \sqrt{\frac{0.25}{90}} ≈ \sqrt{0.00278} ≈ 0.0527\)

Calculating the z-score:

\(z ≈ \frac{0.5778 - 0.5}{0.0527} ≈ \frac{0.0778}{0.0527} ≈ 1.476\)

The p-value is obtained from the standard normal distribution for \(z ≈ 1.476\), which is approximately 0.07.

Conclusion based on p-value

Since the p-value (~0.07) exceeds the common significance level of 0.05, we fail to reject the null hypothesis at the 5% significance level. Therefore, there is insufficient evidence to conclude that college students can distinguish between Sprite and Sierra Mist better than chance in this study.

Calculations of standard deviation for the hypothesis test and confidence interval

  1. Standard deviation of the sample proportion (\(\sigma_{\hat{p}}\)) used in the hypothesis test:
  2. \(\sigma_{\hat{p}} = \sqrt{\frac{p_0 (1 - p_0)}{n}} ≈ 0.0527\)
  3. Standard deviation of \(\hat{p}\) used in confidence interval calculations:
  4. Because the sample estimate \(\hat{p}\) is used, the standard error formula is similar:
  5. \(\text{SE} = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\)
  6. Calculating:
  7. \(\sqrt{\frac{0.5778 \times (1 - 0.5778)}{90}} ≈ \sqrt{\frac{0.5778 \times 0.4222}{90}} ≈ \sqrt{\frac{0.2439}{90}} ≈ \sqrt{0.00271} ≈ 0.0521\)
  8. Reason for using \(\hat{p}\) instead of \(p_0\) in the standard error of the confidence interval:
  9. The standard error uses \(\hat{p}\) because it reflects the observed proportion in the sample, providing an estimate based on actual data rather than an assumed or hypothesized value. Using \(\hat{p}\) makes the confidence interval adaptable and accurately represents the variability in the current sample.

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