Question

Using Data 5.1 on page \(375,\) we find a significant difference in the proportion of fruit flies surviving after 13 days between those eating organic potatoes and those eating conventional (not organic) potatoes. Exercises 6.166 to 6.169 ask you to conduct a hypothesis test using additional data from this study. \(^{40}\) In every case, we are testing $$\begin{array}{ll}H_{0}: & p_{o}=p_{c} \\\H_{a}: & p_{o}>p_{c}\end{array}$$ where \(p_{o}\) and \(p_{c}\) represent the proportion of fruit flies alive at the end of the given time frame of those eating organic food and those eating conventional food, respectively. Also, in every case, we have \(n_{1}=n_{2}=500 .\) Show all remaining details in the test, using a \(5 \%\) significance level. Effect of Organic Raisins after 15 Days After 15 days, 320 of the 500 fruit flies eating organic raisins are still alive, while 300 of the 500 eating conventional raisins are still alive.

Short answer

At the 5% significance level, there is evidence to suggest that a higher proportion of fruit flies are alive at the end of the 15-day period among those eating organic food when compared to those eating conventional food.

Step-by-step solution

State the hypotheses

The null hypothesis is \(H_{0}: p_{o}=p_{c}\) and the alternative hypothesis is \(H_{a}: p_{o}>p_{c}\), where \(p_{o}\) and \(p_{c}\) represent the proportion of fruit flies alive at the end of the given time frame of those eating organic food and those eating conventional food, respectively.

Compute the sample proportions

Compute the sample proportions by dividing the number of successful outcomes (fruit flies still alive) by the total sample size for each group. For the fruit flies eating organic raisins, \(p_{o}=\frac{320}{500}=0.64\). For the fruit flies eating conventional raisins, \(p_{c}=\frac{300}{500}=0.60\).

Compute the standard error

The standard error (SE) is given by the formula \[\sqrt{\frac {(p_{o}(1-p_{o})){n_{o}}}{n} + \frac{(p_{c}(1-p_{c})){n_{c}}}{n}},\] where \(n_{o}=n_{c}=500\) are the sample sizes. Pool the proportions: \(P=\frac{x_{o}+x_{c}}{n_{o}+n_{c}}=\frac{320+300}{1000}=0.62,\] Assumption \( p_{o}=p_{c}\), so \(p_{o}=p_{c}=P=0.62\). Now calculate the SE, \[SE = \sqrt{\frac {(0.62(1-0.62))^2}{n}}, \] \[ =\sqrt{\frac{0.2356}{500}},\] \[=0.0217\].

Compute the z-score

The formula to calculate the z-score is \[Z=\frac{p_{o}-p_{c}}{SE}.\] Substitute the values to get, \[Z=\frac{0.64-0.60}{0.0217}=1.84.\]

Compute the p-value

The test is one-tailed, hence the p-value is the area to the right of the calculated z-value in the standard normal distribution (since the alternative hypothesis suggests that the proportion among flies eating organic food is greater than those eating conventional food). We find that \(P(Z > 1.84)\) is approximately 0.033.

Deciding whether to reject the null hypothesis

A p-value of 0.033 is less than the significance level of 0.05, hence there is sufficient evidence at the 5% level to reject the null hypothesis. This suggests that a higher proportion of fruit flies are alive at the end of the 15-day period among those eating organic food when compared to those eating conventional food.

Key concepts

Null Hypothesis

The null hypothesis is a central concept in hypothesis testing. It is a statement made for the sake of argument that there is no effect or no difference in the context being studied. In simpler terms, it's the default position that there is no association between two measured phenomena.

Using the given exercise as an example, the null hypothesis, denoted as \(H_0\), states that the proportion of surviving fruit flies, denoted by \(p_o\) (eating organic potatoes), is equal to \(p_c\) (eating conventional potatoes). Mathematically, we express this as \(H_0: p_o = p_c\).

It is important to understand that the null hypothesis is what we assume to be true before collecting any data, and hypothesis testing is a method to determine whether the evidence is strong enough to reject this null hypothesis in favor of an alternative hypothesis.

Alternative Hypothesis

The alternative hypothesis, denoted by \(H_a\), is a statement that contradicts the null hypothesis. It is what you aim to support with evidence. For the exercise at hand, the alternative hypothesis claims that the proportion of fruit flies that survive after eating organic potatoes (\(p_o\)) is greater than that of those eating conventional potatoes (\(p_c\)). This is symbolized as \(H_a: p_o > p_c\).

When designing a study or experiment, researchers will try to reject the null hypothesis, thus lending support to the alternative hypothesis. However, they can never fully 'prove' the alternative; they can only gather enough evidence to suggest that it is true.

Significance Level

The significance level, often denoted as \(\alpha\), is a threshold used to judge whether a test statistic is extreme enough to reject the null hypothesis. It's the probability of making the mistake of rejecting a true null hypothesis, also known as a type I error. The lower the significance level, the stronger the evidence required to reject the null hypothesis.

The exercise mentions a 5% significance level, denoted as 0.05, which is common practice in many fields. This means that if the likelihood of observing a test result as extreme as, or more extreme than, the result obtained by chance alone is less than 5%, we would reject the null hypothesis.

Z-score

A z-score is used in hypothesis testing to represent the number of standard deviations a data point is from the mean. It's a way of standardizing the results so that we can understand how unusual they are given a normal distribution.

In the step-by-step solution, the z-score, calculated using the formula \(Z = \frac{p_o - p_c}{SE}\), is 1.84. This represents how far the difference in sample proportions (between fruit flies eating organic versus conventional food) is from the expected difference under the null hypothesis. A high z-score indicates that the observed difference is uncommon under the null hypothesis.

P-value

The p-value measures the probability of obtaining test results at least as extreme as the ones observed during the test, assuming that the null hypothesis is true. In this context, 'extreme' refers to how much the sample data differs from what we would expect if the null hypothesis were correct.

From the exercise result, the p-value is approximately 0.033. Since this p-value is less than the significance level of 0.05, it suggests that the observed data (the survival rates of fruit flies) are unusual enough under the assumption of the null hypothesis that we would reject the null in favor of the alternative hypothesis, indicating a significant difference in the survival rates favoring fruit flies eating organic potatoes.