Question

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

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

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.

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 \(16\% of 556, which equals to 0.16 * 556 = 88.96 \approx 89\)

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 \(p_{males} = 133/556 = 0.24\), \(p_{females} = 89/556 = 0.16\).

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.

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.

Key concepts

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 ≤ 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—our Z-score in this case—and 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.