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Chapter 10: Problem 10

Two TV commercials are developed for marketing a new product. A volunteer test sample of 200 people is randomly split into two groups of 100 each. In a controlled setting, Group A watches commercial A and Group B watches commercial B. In Group A, 25 say they would buy the product. In group \(\mathrm{B}, 20\) say they would buy the product. The marketing manager who devised this experiment concludes that commercial \(\mathrm{A}\) is better. Is this conclusion justified? Analyze the data. If you prefer, use software (such as MINITAB or a web app) for which you can enter summary counts. a. Show all steps of your analysis (perhaps including an appropriate graph) and check assumptions. b. Comment on the manager's conclusion and indicate limitations of the experiment.

Short Answer

Expert verified
The conclusion that commercial A is better may not be justified without statistically significant evidence.

Step by step solution

01

Define Hypotheses

First, we need to define the null and alternative hypotheses for this test: - Null Hypothesis ( H_0 ): There is no difference in the effectiveness of the two commercials, meaning both have the same proportion of people willing to buy. - Alternative Hypothesis ( H_a ): Commercial A is more effective than Commercial B (the proportion of people willing to buy is greater in Group A than Group B).
02

Collect Data and Calculate Observed Proportions

Let's determine the observed proportions of people willing to buy the product from each group:- Group A: \(\hat{p}_A = \frac{25}{100} = 0.25\)- Group B: \(\hat{p}_B = \frac{20}{100} = 0.20\)
03

Test Statistic

To compare the proportions, we use a hypothesis test for two proportions. The test statistic is calculated as follows:\[z = \frac{\hat{p}_A - \hat{p}_B}{\sqrt{\hat{p} \cdot (1 - \hat{p}) \cdot (\frac{1}{n_A} + \frac{1}{n_B})}}\]Where \(\hat{p} = \frac{x_A + x_B}{n_A + n_B} = \frac{25 + 20}{100 + 100} = 0.225\). Plug these values into the formula to calculate \(z\).
04

Compute P-Value

Using the calculated \(z\)-value, determine the p-value, which corresponds to the probability of observing a test statistic as extreme as, or more extreme than, the observed \(z\)-value under the null hypothesis. This usually involves looking up the \(z\)-value in a standard normal distribution table or using statistical software.
05

Draw Conclusions

According to the p-value and significance level (typically \(\alpha = 0.05\)), decide whether to reject or fail to reject the null hypothesis. If the p-value is less than \(\alpha\), reject the null hypothesis.
06

Limitations

Consider the limitations of the experiment. Random assignment and sample size can affect the reliability. Additionally, other factors such as unaccounted confounding variables or biases might be present in the controlled environment that could influence the likelihood of purchase.

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Key Concepts

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

Null and Alternative Hypotheses
In the realm of statistical hypothesis testing, defining clear hypotheses is crucial. The null hypothesis, denoted as \( H_0 \), in this context is that there is no difference between the two TV commercials in terms of their effectiveness in convincing viewers to buy the product. Essentially, \( H_0 \) asserts that the proportion of people willing to buy the product is the same for both commercials. On the flip side, the alternative hypothesis, \( H_a \), suggests that Commercial A is more effective, implying that a higher proportion of viewers for Commercial A would purchase the product compared to Commercial B. Formulating these hypotheses helps set the stage for subsequent statistical analysis and comparison.
Proportion Comparison
Proportion comparison is key to determining the effectiveness of the commercials. To assess this, we calculate the observed proportions from each group. For Group A, the proportion \( \hat{p}_A \) is calculated as \( \frac{25}{100} = 0.25 \). For Group B, the observed proportion \( \hat{p}_B \) is \( \frac{20}{100} = 0.20 \). These values indicate that in Group A, 25% of viewers are willing to buy the product, while only 20% in Group B show the same willingness.

By comparing these two proportions, we attempt to infer whether the difference is statistically significant or just due to random chance. A statistical test for two proportions aids in discerning this, by calculating a test statistic based on the observed differences.
P-Value Calculation
The p-value is a critical component in statistical hypothesis testing. It quantifies the evidence against the null hypothesis. After calculating the test statistic, commonly a \( z \)-value in a test for two proportions, the p-value is determined as the probability of observing a test statistic as extreme as the observed \( z \)-value, assuming the null hypothesis is true.

To locate this p-value, the \( z \)-value is referenced against a standard normal distribution, either using statistical software or a standard normal table. If the p-value is smaller than a chosen significance level, usually \( \alpha = 0.05 \), it suggests that rejecting the null hypothesis is justified, indicating a statistically significant difference between the commercials.
Experiment Limitations
Understanding the limitations of an experiment clarifies the bounds of its conclusions. For the TV commercial test, several factors limit the reliability of the results. Firstly, the sample size. With just 100 people per group, the results could be influenced significantly by randomness. Also, the experiment's controlled settings might not reflect real-world conditions, affecting how people would normally respond.

Another concern is the possibility of confounding variables. These are factors not accounted for that could affect the outcome, such as prior consumer attitudes or demographics. Moreover, biases in assignment or measurement can skew results, undermining the conclusions drawn. By acknowledging these limitations, one can better assess the generalizability and potential bias in the experiment's findings.

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Most popular questions from this chapter

From the box formula for the standard error at the end of Section 10.1 \(\frac{s e(\text { estimate } 1-\text { estimate } 2)=}{\sqrt{[s e(\text { estimate } 1)]^{2}+[s e(\text { estimate } 2)]^{2}}}\) if you know the se for each of two independent estimates, you can find the se of their difference. This is useful because often articles report an se for each sample mean or proportion, but not the se or a confidence interval for their difference. Many medical studies have used a large sample of subjects from Framingham, Massachusetts, who have been followed since 1948. A study (Annual of Internal Medicine, vol. \(138,2003,\) pp. \(24-32\) ) estimated the number of years of life lost by being obese and a smoker. For females of age 40 , adjusting for other factors, the number of years of life left were estimated to have a mean of \(46.3(s e=0.6)\) for nonsmokers of normal weight and a mean of \(33.0(s e=1.8)\) for smokers who were obese. Construct a \(95 \%\) confidence interval for the population mean number of years lost. Interpret.

Multiple choice: Sample size and significance If the sample proportions in Example 4 comparing cancer death rates for aspirin and placebo had sample sizes of only 1000 each, rather than about 11,000 each, then the \(95 \%\) confidence interval for \(\left(p_{1}-p_{2}\right)\) would be (-0.007,0.021) rather than \((0.003,0.011) .\) This reflects that a. When an effect is small, it may take very large samples to have much power for establishing statistical significance. b. Smaller sample sizes are preferable because there is more of a chance of capturing 0 in a confidence interval. c. Confidence intervals get wider when sample sizes get larger. d. The confidence interval based on small sample sizes must be in error because it is impossible for the parameter to take a negative value.

Empagliflozin and renal function over time A study published in June 2016 in New England Journal of Medicine wanted to determine the long-term renal effects (measured by eGFR: estimated glomerular filtration rate) of empagliflozin in patients with type 2 Diabetes. 7020 patients with type 2 Diabetes at 590 sites in 42 countries received at least one dose of a study drug. Patients were randomly assigned to receive either empagliflozin (at a dose level of either \(10 \mathrm{mg}\) or \(25 \mathrm{mg}\) ) or a placebo once daily in addition to standard care. The difference between the study groups in the average rate of change in eGFR was estimated after a duration of 4 weeks. a. The authors stated that there was a short-term decrease in the eGFR in the empagliflozin groups, with \(95 \%\) confidence interval of weekly decreases of \(0.62 \pm 0.04 \mathrm{ml}\) per minute per \(1.73 \mathrm{~m}^{2}\) of body- surface area in the 10-mg group. Interpret the confidence interval. b. The authors also provided a p-value that is \(<0.001\) for the comparisons in eGFR means of the \(10 \mathrm{mg}\) dose \(\mathrm{em}\) pagliflozin group with the placebo group. Specify the hypotheses for this test of comparison of means, which was two-sided. Interpret the p-value obtained.

Vegetarians more liberal? When a sample of social science graduate students at the University of Florida gave their responses on political ideology (ranging from \(1=\) very liberal to \(7=\) very conservative \(),\) the mean was \(3.18(s=1.72)\) for the 51 nonvegetarian students and \(2.22(s=0.67)\) for the 9 vegetarian students. Software for comparing the means provides the printout, which shows results first for inferences that assume equal population standard deviations and then for inferences that allow them to be unequal. \(\begin{array}{ccclc}\text { Sample } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } \\ 1 & 51 & 3.18 & 1.72 & 0.24 \\ 2 & 9 & 2.220 & 0.670 & 0.22 \\ \text { Difference } & =\mu(1)-\mu(2) & & \end{array}\) Estimate for difference: 0.960 95?a CI for difference: (-0.210,2.130) T-Test of difference \(=0\) (vs \(\neq):\) T-Value \(=1.64\) \(\mathrm{P}\) -Value \(=0.106 \mathrm{DF}=58\) Both use Pooled StDev \(=1.6162\) 95? CI for difference: (0.289,1.631) T-Test of difference \(=0(\mathrm{vs} \neq):\) T-Value \(=2.92\) P-Value \(=0.007 \quad \mathrm{DF}=30\) a. Explain why the results of the two approaches differ so much. Which do you think is more reliable? b. State your conclusion about whether the true means are plausibly equal.

c. A \(95 \%\) confidence interval for the difference between the mean level of tryptophan in the treated versus the control group is \((-0.007,0.004) .\) Interpret. Geneticists hypothesized that one reason for a prolonged lifespan observed in yeast cells treated with the common drug ibuprofen stems from inhibiting the uptake of certain amino acids. In an experiment, they compared levels of amino acids in cells treated with ibuprofen to cells from an untreated control group. For the amino acid tryptophan, the six measurements in the treated group had a mean level of 0.0067 with a standard deviation of 0.0038 . The six measurements in the control group had a mean level of 0.0080 with a standard deviation of 0.0050 . a. Identify the response variable and explanatory variable. Indicate whether each is quantitative or categorical. b. Find the standard error of the sample mean in each group and the standard error of the difference in the sample means.
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