Question

A researcher wishes to compare the prices for randomly selected prescription drugs in the United States with those in Canada. The same drugs and dosages were compared in each country. At \(\alpha=0.05,\) can it be concluded that the drugs in Canada are cheaper? $$ \begin{array}{l|cccccc} \text { Drug } & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text { United States } & 3.31 & 2.27 & 2.54 & 3.13 & 23.40 & 3.16 \\ \hline \text { Canada } & 1.47 & 1.07 & 1.34 & 1.34 & 21.44 & 1.47 \end{array} $$ $$ \begin{array}{l|cccc} \text { Drug } & 7 & 8 & 9 & 10 \\ \hline \text { United States } & 1.98 & 5.27 & 1.96 & 1.11 \\ \hline \text { Canada } & 1.07 & 3.39 & 2.22 & 1.13 \end{array} $$

Short answer

The drugs in Canada are cheaper at \( \alpha=0.05 \) significance level.

Step-by-step solution

Formulate Null and Alternative Hypotheses

The null hypothesis \( H_0 \) is that the average price of prescription drugs in Canada is equal to that in the United States, which means \( \mu_{d} = 0 \), where \( \mu_{d} \) is the average difference in prices. The alternative hypothesis \( H_a \) is that the average price of prescription drugs in Canada is less than in the United States, indicated by \( \mu_{d} > 0 \).

Calculate Mean and Standard Deviation of Differences

Compute the difference in prices for each drug and find the mean \( \bar{d} \) and standard deviation \( s_d \) of these differences. The differences are: 1.84, 1.20, 1.20, 1.79, 1.96, 1.69, 0.91, 1.88, -0.26, -0.02. Calculate: \[ \bar{d} = \frac{\sum{d}}{n} = \frac{11.19}{10} = 1.119, \] \[ s_d = \sqrt{\frac{\sum{(d_i - \bar{d})^2}}{n-1}}, \] with \( n = 10 \).

Conduct t-Test for Paired Samples

Using the t-test formula for paired samples: \[ t = \frac{\bar{d}}{s_d / \sqrt{n}}, \] calculate \( t \) value. First, calculate \( s_d = 0.7398 \) (using the calculated differences earlier), then \[ t = \frac{1.119}{0.7398 / \sqrt{10}} = 4.7747. \]

Determine Critical t-Value and Compare

Look up the critical t-value from the t-distribution table for \( \alpha = 0.05 \) and \( df = n-1 \). For \( df = 9 \), the critical t-value is approximately 1.833. Since our calculated t-value \( 4.7747 \) is greater than the critical value, we reject the null hypothesis.

Conclusion

Since the t-test result is significant, we have enough evidence to conclude that, at the 0.05 significance level, the prescription drugs in Canada are cheaper than those in the United States.

Key concepts

Paired t-test

A paired t-test examines the differences between two related groups of data. In this case, we're comparing the prices of the same prescription drugs in the United States and Canada. The key here is that the same drugs are paired together in each location.

This test checks whether the average difference between these paired observations is significantly different from zero.

To perform a paired t-test, you follow these steps:

• First, calculate the difference between each pair of observations.
• Next, determine the mean and standard deviation of these differences.
• Finally, use these values to compute the t-statistic and compare it to the critical t-value.

By applying this test, we determine if there is a significant price difference between the US and Canada, rather than relying on arbitrary judgment.

Null and Alternative Hypotheses

The hypotheses frame your research question into tests you can statistically evaluate. In hypothesis testing, we set a null hypothesis (H_0) and an alternative hypothesis (H_a).

For our exercise, the null hypothesis suggests there's no difference in average drug prices between the US and Canada, mathematically expressed as \( \mu_{d} = 0 \). This means the mean difference in prices, \( \mu_{d} \), is zero.
The alternative hypothesis assumes that drug prices in Canada are cheaper, implying \( \mu_{d} > 0 \).
The test's result will either reject or fail to reject the null hypothesis based on statistical evidence.

• If the null hypothesis is rejected, it suggests significant evidence for the alternative hypothesis.
• If we fail to reject the null hypothesis, it indicates insufficient evidence of a difference.

Understanding these hypotheses is crucial as they lay the groundwork for identifying if a statistically significant difference exists.

Significance Level

The significance level, often represented as \( \alpha \), determines the probability of rejecting the null hypothesis when it's actually true. For this exercise, the chosen level is 0.05.

This means we're willing to accept a 5% chance of making a Type I error — claiming a difference exists when there isn't one.
Using the significance level, we determine the critical t-value from the t-distribution table, which acts as a threshold.
In hypothesis testing:

• If our calculated t-statistic exceeds this critical value, we reject the null hypothesis.
• If it doesn't, we fail to reject the null hypothesis.

Choosing an appropriate significance level is important, as it dictates the strictness of the test and its conclusions. In this case, the test revealed enough evidence to conclude that Canadian drug prices are indeed lower at the 0.05 level.