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Female Models In a study to assess various effects of using a female model in automobile advertising, 100 men were shown photographs of two automobiles matched for price, color, and size, but of different makes. One of the automobiles was shown with a female model to 50 of the men (group \(\mathrm{A}\) ), and both automobiles were shown without the model to th other 50 men (group B). In group A, the automobile shown with the model was judged as more expensive by 37 men; in group \(B\), the same automobile was judged as the more expensive by 23 men. Do these results indicate that using a female model influences the perceived cost of an automobile? Use a one-tailec test with \(\alpha=.05 .\)

Short Answer

Expert verified
Answer: Yes, based on the analysis, using a female model in automobile advertising does significantly influence the perceived cost of an automobile among men.

Step by step solution

01

State the null and alternative hypotheses

The null hypothesis \((H_0)\) states that there's no difference between the proportions of men who judge the automobile as more expensive in both groups. The alternative hypothesis \((H_a)\) states that the proportion of men who judge the automobile as more expensive is significantly higher in Group A (with the model). \(H_0: p_A = p_B\) \(H_a: p_A > p_B\)
02

Calculate the sample proportions

In Group A, 37 out of 50 men judged the automobile with the model as more expensive. In Group B, 23 out of 50 men judged the same automobile as more expensive. \(p_A = \frac{37}{50} = 0.74\) \(p_B = \frac{23}{50} = 0.46\)
03

Calculate the pooled proportion and standard error

The pooled proportion \((p)\) is calculated by dividing the total number of men who judged the automobile as more expensive in both groups by the total number of men. \(p = \frac{37+23}{50+50} = \frac{60}{100} = 0.60\) The standard error (SE) for the difference in proportions is calculated as follows: \(SE = \sqrt{\frac{p(1-p)}{n_A} + \frac{p(1-p)}{n_B}} = \sqrt{\frac{0.60(1-0.60)}{50} + \frac{0.60(1-0.60)}{50}} = 0.1225\)
04

Calculate the test statistic

The test statistic (z) is calculated using the formula: \(z = \frac{(p_A - p_B) - 0}{SE} = \frac{(0.74 - 0.46) - 0}{0.1225} = 2.285\)
05

Determine the p-value from the z-score

Using the z-score calculator or z-table, we can find the p-value (the probability of getting a z-score of 2.285 or higher). The p-value is found to be approximately 0.011.
06

Compare the p-value with the significance level

Compare the p-value (0.011) with the significance level \(\alpha = 0.05\). Since the p-value is less than the significance level, we reject the null hypothesis.
07

Interpret the results

Based on the analysis, we reject the null hypothesis that there is no difference between the proportions of men who judge the automobile as more expensive in both groups. The results suggest that using a female model in automobile advertising does significantly influence the perceived cost of an automobile.

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Key Concepts

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

Null and Alternative Hypothesis
Hypothesis testing begins with the establishment of the null and alternative hypotheses. The null hypothesis ( H_0) represents a statement of no effect or no difference. It serves as the default assumption that there is no relationship between the variables under investigation. In our automobile advertising study, the null hypothesis asserts that the presence of a female model does not impact the perceived cost of the automobile.

The alternative hypothesis ( H_a), on the other hand, challenges the null hypothesis by proposing that there is an effect or a difference. For the given exercise, the alternative hypothesis posits that the automobile shown with a female model is judged as more expensive by a higher proportion of men than the same automobile shown without it. It's crucial to define these hypotheses precisely, as they directly influence the direction and purpose of the statistical test to be performed.
Pooled Proportion
The pooled proportion is a weighted average of individual sample proportions when we assume under the null hypothesis that the two populations have the same proportion. It combines the successes from both groups and divides by the total number of observations.

In the context of our automobile advertising study, we find the pooled proportion by adding the number of men who perceived the automobile as more expensive in both groups and then dividing by the total number of men. This pooled proportion represents what the proportion would look like if there truly were no difference—that is, if the null hypothesis were true.
Standard Error
The standard error quantifies the variability of the sampling distribution for the difference in sample proportions. It's an estimate of the standard deviation of the sampling distribution and it helps determine how far the sample statistic is likely to be from the population parameter.

The standard error in the automobile study is calculated under the assumption that the null hypothesis is correct. It reflects how much we expect the difference in the proportions ( p_A - p_B) to vary due to random sampling. If our calculated test statistic falls far outside the range that the standard error suggests is likely, it presents evidence against the null hypothesis.
Test Statistic
A test statistic is a standardized value that is calculated from sample data during a hypothesis test. It is used to decide whether to reject the null hypothesis. The formula for the test statistic will vary depending on the statistical test being performed.

In this case, we used the test statistic ( z) to assess the significance of the difference between the two sample proportions. We calculated the z-value by subtracting the hypothesized difference (under the null hypothesis) from the observed difference ( p_A - p_B) and then dividing by the standard error of the difference. The resulting z-value tells us how many standard errors the observed difference is away from the hypothesized value.
P-Value
The p-value plays a central role in hypothesis testing—it’s the probability of obtaining test results at least as extreme as the results observed during the test, assuming that the null hypothesis is true. A small p-value indicates that the observed data is unlikely under the null hypothesis, leading us to consider rejecting the null hypothesis.

In the exercise, the p-value obtained from the z-score indicates the probability of observing such a marked difference in perceptions if the null hypothesis were true. If this p-value is smaller than our predetermined significance level ( α), we have significant evidence to doubt the null hypothesis and consider the alternative hypothesis as a better explanation of the observed data.

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Most popular questions from this chapter

Sweet Potato Whitefly Suppose that \(10 \%\) of the fields in a given agricultural area are infested with the sweet potato whitefly. One hundred fields in this area are randomly selected, and 25 are found to be infested with whitefly. a. Assuming that the experiment satisfies the conditions of the binomial experiment, do the data indicate that the proportion of infested fields is greater than expected? Use the \(p\) -value approach, and test using a \(5 \%\) significance level. b. If the proportion of infested fields is found to be significantly greater than \(.10,\) why is this of practical significance to the agronomist? What practical conclusions might she draw from the results?

Generation Next Born between 1980 and 1990, Generation Next was the topic of Exercise \(8.60 .^{16}\) In a survey of 500 female and 500 male students in Generation Next, 345 of the females and 365 of the males reported that they decided to attend college in order to make more money. a. Is there a significant difference in the population proportions of female and male students who decided to attend college in order to make more money? Use \(\alpha=.01\). b. Can you think of any reason why a statistically significant difference in these population proportions might be of practical importance? To whom might this difference be important?

Plant Genetics A peony plant with red petals was crossed with another plant having streaky petals. A geneticist states that \(75 \%\) of the offspring resulting from this cross will have red flowers. To test this claim, 100 seeds from this cross were collected and germinated, and 58 plants had red petals. a. What hypothesis should you use to test the geneticist's claim? b. Calculate the test statistic and its \(p\) -value. Use the \(p\) -value to evaluate the statistical significance of the results at the \(1 \%\) level.

Independent random samples of 80 measurements were drawn from two quantitative populations, 1 and \(2 .\) Here is a summary of the sample data: $$\begin{array}{lcc} & \text { Sample 1 } & \text { Sample 2 } \\\\\hline \text { Sample Size } & 80 & 80 \\\\\text { Sample Mean } & 11.6 & 9.7 \\\\\text { Sample Variance } & 27.9 & 38.4\end{array}$$ a. If your research objective is to show that \(\mu_{1}\) is larger than \(\mu_{2}\), state the alternative and the null hypotheses that you would choose for a statistical test. b. Is the test in part a one- or two-tailed? c. Calculate the test statistic that you would use for the test in part a. Based on your knowledge of the standard normal distribution, is this a likely or unlikely observation, assuming that \(H_{0}\) is true and the two population means are the same? d. \(p\) -value approach: Find the \(p\) -value for the test. Test for a significant difference in the population means at the \(1 \%\) significance level. e. Critical value approach: Find the rejection region when \(\alpha=.01 .\) Do the data provide sufficient evidence to indicate a difference in the population anc?

a. Define \(\alpha\) and \(\beta\) for a statistical test of hypothesis. b. For a fixed sample size \(n\), if the value of \(\alpha\) is decreased, what is the effect on \(\beta\) ? c. In order to decrease both \(\alpha\) and \(\beta\) for a particular alternative value of \(\mu,\) how must the sample size change?

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