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Suppose the manufacturer of Advil, a common headache remedy, recently developed a new formulation of the drug that is claimed to be more effective. To evaluate the new drug, a sample of 200 current users is asked to try it. After a one-month trial, 180 indicated the new drug was more effective in relieving a headache. At the same time a sample of 300 current Advil users is given the current drug but told it is the new formulation. From this group, 261 said it was an improvement. At the .05 significance level can we conclude that the new drug is more effective?

Short Answer

Expert verified
The new drug is more effective at a 0.05 significance level.

Step by step solution

01

Set up Hypotheses

We begin by setting up our null and alternative hypotheses. The null hypothesis, \( H_0 \), states that the proportion of people who found the new drug effective is less than or equal to the proportion who found the current drug effective. The alternative hypothesis, \( H_1 \), states that the proportion of people who found the new drug effective is greater than the proportion who found the current drug effective. Mathematically, these are:- \( H_0: p_1 \leq p_2 \)- \( H_1: p_1 > p_2 \)where \( p_1 \) is the proportion for the new drug and \( p_2 \) is the proportion for the current drug.
02

Calculate Sample Proportions

Calculate the sample proportions for both groups. For the new drug users:\[ \hat{p}_1 = \frac{180}{200} = 0.9 \]For the current drug users (placebo group):\[ \hat{p}_2 = \frac{261}{300} = 0.87 \]
03

Calculate Pooled Proportion

The pooled proportion, \( \hat{p} \), combines the successes and total sample sizes of both groups:\[ \hat{p} = \frac{180 + 261}{200 + 300} = \frac{441}{500} = 0.882 \]
04

Compute Standard Error of the Difference

Calculate the standard error (SE) of the difference between the two sample proportions using the pooled proportion:\[ SE = \sqrt{\hat{p}(1-\hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right) } = \sqrt{0.882(1-0.882) \left( \frac{1}{200} + \frac{1}{300} \right) } \]
05

Calculate Test Statistic

Using the sample proportions, pooled proportion, and standard error calculated, we find the test statistic, \( z \):\[ z = \frac{\hat{p}_1 - \hat{p}_2}{SE} \] Insert the numbers computed above into this formula and solve for \( z \).
06

Determine Critical Value and Decision

For a significance level of \( \alpha = 0.05 \) and a one-tailed test, find the critical value for \( z \) using a standard normal distribution table (approx. 1.645). Compare the calculated \( z \) value with the critical value. If \( z > 1.645 \), reject the null hypothesis.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis is the starting assumption that there is no effect or no difference between groups. It's the position you begin with, and it proposes that any observed variation is due to chance. For instance, in the experiment involving Advil's newly formulated drug, the null hypothesis (\( H_0 \)) states that the proportion of people who found the new drug effective (\( p_1 \)) is less than or equal to the proportion of people who found the existing drug effective (\( p_2 \)). In simpler words, we assumed initially that the new drug has a similar or lower effectiveness than the current one. This hypothesis acts as a neutral benchmark to compare the outcome of the research.

To summarize, the null hypothesis is all about preserving the status quo until enough evidence suggests a change is warranted. It's crucial because rejecting or failing to reject it directly informs the conclusions about the effectiveness of the treatment under study.
Alternative Hypothesis
The alternative hypothesis is central to hypothesis testing because it represents what the researcher aims to show. In contrast to the null hypothesis, the alternative hypothesis proposes that there is an effect or a difference. For the Advil experiment, the alternative hypothesis (\( H_1 \)) posits that the proportion (\( p_1 \)) of people who find the new drug effective is greater than the proportion (\( p_2 \)) for the current drug. This hypothesis suggests that the new formulation is indeed more effective.

Clearly defining the alternative hypothesis is crucial because it sets the direction for the conclusions. It answers what researchers genuinely want to find out. When we undertake hypothesis testing, we gather data to provide evidence against the null hypothesis, and in support of the alternative hypothesis.
Sample Proportions
Sample proportions give a snapshot of the observed phenomena in a sample. They are calculated by dividing the number of positive outcomes by the total number in the sample. For comparing drug effectiveness, sample proportions are critical. In the Advil study, the sample proportion for the new drug group (\( \hat{p}_1 \)) was calculated as:
  • 180 people found it effective out of 200, resulting in \( \hat{p}_1 = \frac{180}{200} = 0.9 \).
Similarly, for the current drug users (placebo group):
  • 261 users out of 300 found it effective, giving \( \hat{p}_2 = \frac{261}{300} = 0.87 \).

These proportions are vital for hypothesis testing as they give us the observed data used in further calculations, such as determining pooled proportions and calculating standard errors. This allows us to evaluate the difference in effectiveness between the groups.
Statistical Significance
Statistical significance is a cornerstone concept in hypothesis testing. It helps determine whether the difference observed between groups (like the effectiveness between Advil's new and current drugs) is likely due to chance or a real effect. Significance is assessed using a p-value and a pre-chosen level of significance (alpha, \( \alpha \)).

In the Advil example, the alpha was set at 0.05. A calculated test statistic (z-value) is compared against a critical value from standard normal distribution tables. For a one-tailed test at 0.05 significance, the critical z-value is approximately 1.645. If the computed z is greater than 1.645, then the null hypothesis is rejected, suggesting a statistically significant difference.

Thus, statistical significance gives us the power to make informed decisions based on data analysis. It doesn't guarantee practical significance, but it allows researchers to conclude whether the observed effect is strong enough not to be attributed to random chance.

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Most popular questions from this chapter

Fry Brothers Heating and Air Conditioning, Inc. employs Larry Clark and George Murnen to make service calls to repair furnaces and air conditioning units in homes. Tom Fry, the owner, would like to know whether there is a difference in the mean number of service calls they make per day. Assume the population standard deviation for Larry Clark is 1.05 calls per day and 1.23 calls per day for George Murnen. A random sample of 40 days last year showed that Larry Clark made an average of 4.77 calls per day. For a sample of 50 days George Murnen made an average of 5.02 calls per day. At the .05 significance level, is there a difference in the mean number of calls per day between the two employees? What is the \(p\) -value?

A sample of 40 observations is selected from one population with a population standard deviation of \(5 .\) The sample mean is \(102 .\) A sample of 50 observations is selected from a second population with a population standard deviation of \(6 .\) The sample mean is \(99 .\) Conduct the following test of hypothesis using the .04 significance level. $$ \begin{array}{l} H_{0}: \mu_{1}=\mu_{2} \\ H_{1}: \mu_{1} \neq \mu_{2} \end{array} $$ a. Is this a one-tailed or a two-tailed test? b. State the decision rule. c. Compute the value of the test statistic. d. What is your decision regarding \(H_{0}\) e. What is the \(p\) -value?

Each month the National Association of Purchasing Managers publishes the NAPM index. One of the questions asked on the survey to purchasing agents is: Do you think the economy is expanding? Last month, of the 300 responses, 160 answered yes to the question. This month, 170 of the 290 responses indicated they felt the economy was expanding. At the .05 significance level, can we conclude that a larger proportion of the agents believe the economy is expanding this month?

Mary Jo Fitzpatrick is the vice president for Nursing Services at St. Luke's Memorial Hospital. Recently she noticed in the job postings for nurses that those that are unionized seem to offer higher wages. She decided to investigate and gathered the following information. $$ \begin{array}{|lccc|} \hline & & \text { Population } & \\ \text { Group } & \text { Mean Wage } & \text { Standard Deviation } & \text { Sample Size } \\ \hline \text { Union } & \$ 20.75 & \$ 2.25 & 40 \\ \text { Nonunion } & \$ 19.80 & \$ 1.90 & 45 \\ \hline \end{array} $$ Would it be reasonable for her to conclude that union nurses earn more? Use the .02 significance level. What is the \(p\) -value?

Clark Heter is an industrial engineer at Lyons Products. He would like to determine whether there are more units produced on the night shift than on the day shift. Assume the population standard deviation for the number of units produced on the day shift is 21 and is 28 on the night shift. A sample of 54 day-shift workers showed that the mean number of units produced was 345\. A sample of 60 night-shift workers showed that the mean number of units produced was \(351 .\) At the .05 significance level, is the number of units produced on the night shift larger?

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