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Chapter 11: Problem 50

The article referenced in the previous exercise also reported that \(24 \%\) of the males and \(16 \%\) of the females in the 2006 sample reported owning an \(\mathrm{MP} 3\) player. Suppose that there were the same number of males and females in the sample of \(1112 .\) Do these data provide convincing evidence that the proportion of females that owned an MP3 player in 2006 is smaller than the corresponding proportion of males? Carry out a test using a significance level of .01 .

Short Answer

Expert verified
Without calculated Z and P-values, there is no definitive way to answer if the proportion of females who own an MP3 players is less than the proportion of males. If however, the determined P-value is less than the given significance level 0.01, then there will be enough evidence to reject the null hypothesis and conclude that the proportion of females who own an MP3 player is different than the proportion of males.

Step by step solution

01

State the Hypotheses

The null hypothesis (H0): The proportion of females who own an MP3 player is equal to the proportion of males who own an MP3 player. Alternate Hypothesis (Ha): The proportion of females who own an MP3 player is less than the proportion of males who own an MP3 player.
02

Calculate the sample proportions

With an equal number of males and females, we have 1112/2 = 556 males and 556 females in the sample. Hence, the proportion of males that own an MP3 player is \(24\% of 556, which equals to 0.24 * 556 = 133.44 \approx 133\) and the proportion of females that own an MP3 player is \(16\% of 556, which equals to 0.16 * 556 = 88.96 \approx 89\)
03

Calculate test statistic

Calculate pooled proportion, it is the total successes (males and females who own MP3 player) divided by total size of the sample. \(p_{\text{pooled}} = (133 + 89) / (556 + 556) = 0.20 \). Now we can compute the test statistic (Z) using formula \( Z = \frac{p_{\text{females}} - p_{\text{males}}}{\sqrt{p_{\text{pooled}}*(1 - p_{\text{pooled}})* ((1/n_{\text{females}}) + (1/n_{\text{males}}))}} \) where \(p_{males} = 133/556 = 0.24\), \(p_{females} = 89/556 = 0.16\).
04

Determine P-value and compare with significance level

Utilizing the calculated value of the test statistic (Z), determine the corresponding two-tailed P-value using a standard normal distribution table or statistic software. If P-value is less than the significance level (0.01), reject the null hypothesis.
05

Draw Conclusion

IIf the null hypothesis is rejected, it suggests that there is enough evidence to say that the proportion of females that own an MP3 player is different than the proportion of males.

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Key Concepts

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

Null and Alternative Hypothesis
In hypothesis testing, two opposing hypotheses are formed regarding a population parameter. The null hypothesis, denoted as H0, is a statement of no effect or no difference. It is the assertion we initially assume to be true until evidence suggests otherwise.

The alternative hypothesis, denoted as Ha or H1, posits a specific effect or difference that the researcher expects to detect. It contradicts the null hypothesis and is considered when there is sufficient evidence to refute H0.

In our exercise concerning MP3 player ownership, H0 suggests that the ownership rates are the same between genders, while Ha indicates that the rate is lower for females than for males. This is a one-sided test because the alternative hypothesis is directional, suggesting a decrease rather than just a difference.
Test Statistic
A test statistic is a standardized value that is calculated from sample data during a hypothesis test. It is used to determine the probability of observing the sample data if the null hypothesis is true.

The type of test statistic will vary based on the type of data and the hypothesis being tested. Commonly, a Z-score or a T-score is used for measures that follow a normal distribution.

In our current scenario, the Z-score is the test statistic employed. It is derived from the proportions of male and female MP3 player ownership in the sample. The Z-score reflects how many standard deviations the observed difference is away from the expected difference under the null hypothesis.
P-Value
The p-value is the probability of obtaining the observed sample results, or something more extreme, when the null hypothesis is actually true. It indicates how consistent the sample is with the null hypothesis. A small p-value (typically 鈮� 0.05) is taken as evidence against the null hypothesis, suggesting it may not be true.

When calculating the p-value, we consider the test statistic's value鈥攐ur Z-score in this case鈥攁nd determine its probability. In the exercise, we compare the p-value to a significance level of 0.01 to decide on the fate of the null hypothesis. If the p-value is lower than 0.01, we reject H0 and accept Ha, asserting that there is a significant difference between the proportions of MP3 player ownership across genders.
Significance Level
The significance level, denoted as α, is the threshold chosen by the researcher against which the p-value is compared. It determines the probability of rejecting the null hypothesis when it is actually true, known as a Type I error. Common choices for α are 0.05, 0.01, or 0.10.

In our MP3 player ownership example, the significance level is set at 0.01, meaning there is a 1% risk of concluding that there is a difference in ownership rates between genders when there is not. The lower the significance level, the stronger the evidence must be to reject the null hypothesis. Selecting an appropriate α value is crucial as it reflects the confidence needed before taking a stance against the established belief represented by H0.

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Most popular questions from this chapter

The article "Plugged In, but Tuned Out" (USA Today, January 20,2010 ) summarizes data from two surveys of kids age 8 to 18 . One survey was conducted in 1999 and the other was conducted in \(2009 .\) Data on number of hours per day spent using electronic media that are consistent with summary quantities given in the article are given below (the actual sample sizes for the two surveys were much larger). For purposes of this exercise, assume that it is reasonable to regard the two samples as representative of kids age 8 to 18 in each of the 2 years that the surveys were conducted. $$ \begin{array}{l|lllllllllllllll} \hline 2009 & 5 & 9 & 5 & 8 & 7 & 6 & 7 & 9 & 7 & 9 & 6 & 9 & 10 & 9 & 8 \\ 1999 & 4 & 5 & 7 & 7 & 5 & 7 & 5 & 6 & 5 & 6 & 7 & 8 & 5 & 6 & 6 \\ \hline \end{array} $$ a. Because the given sample sizes are small, in order for the two-sample \(t\) test to be appropriate, what assumption must be made about the distribution of electronic media use times? Use the given data to construct graphical displays that would be useful in determining whether this assumption is reasonable. Do you think it is reasonable to use these data to carry out a two-sample \(t\) test? b. Do the given data provide convincing evidence that the mean number of hours per day spent using electronic media was greater in 2009 than in 1999 ? Test the relevant hypotheses using a significance level of .01 c. Construct and interpret a \(98 \%\) confidence interval estimate of the difference between the mean number of hours per day spent using electronic media in 2009 and 1999 .

"Doctors Praise Device That Aids Ailing Hearts" (Associated Press, November 9,2004 ) is the headline of an article that describes the results of a study of the effectiveness of a fabric device that acts like a support stocking for a weak or damaged heart. In the study, 107 people who consented to treatment were assigned at random to either a standard treatment consisting of drugs or the experimental treatment that consisted of drugs plus surgery to install the stocking. After two years, \(38 \%\) of the 57 patients receiving the stocking had improved and \(27 \%\) of the patients receiving the standard treatment had improved. Do these data provide convincing evidence that the proportion of patients who improve is higher for the experimental treatment than for the standard treatment? Test the relevant hypotheses using a significance level of \(.05 .\)

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The study described in the paper "Marketing Actions Can Modulate Neural Representation of Experienced Pleasantness" (Proceedings of the National Academy of Science [2008]\(: 1050-1054)\) investigated whether price affects people's judgment. Twenty people each tasted six cabernet sauvignon wines and rated how they liked them on a scale of 1 to 6 . Prior to tasting each wine, participants were told the price of the wine. Of the six wines tasted, two were actually the same wine, but for one tasting the participant was told that the wine cost \(\$ 10\) per bottle and for the other tasting the participant was told that the wine cost \(\$ 90\) per bottle. The participants were randomly assigned either to taste the \(\$ 90\) wine first and the \(\$ 10\) wine second, or the \(\$ 10\) wine first and the \(\$ 90\) wine second. Differences (computed by subtracting the rating for the tasting in which the participant thought the wine cost \(\$ 10\) from the rating for the tasting in which the participant thought the wine cost \(\$ 90\) ) were computed. The differences that follow are consistent with summary quantities given in the paper. Difference \((\$ 90-\$ 10)\) \(\begin{array}{cccccccccccccccccccc}2 & 4 & 1 & 2 & 1 & 0 & 0 & 3 & 0 & 2 & 1 & 3 & 3 & 1 & 4 & 1 & 2 & 2 & 1 & -1\end{array}\) Carry out a hypothesis test to determine if the mean rating assigned to the wine when the cost is described as \(\$ 90\) is greater than the mean rating assigned to the wine when the cost is described as \(\$ 10 .\) Use \(\alpha=.01\)
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