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Marketing companies have collected data implying that teenage girls use more ring tones on their cellular phones than teenage boys do. In one particular study of 40 randomly chosen teenage girls and boys (20 of each) with cellular phones, the mean number of ring tones for the girls was 3.2 with a standard deviation of 1.5. The mean for the boys was 1.7 with a standard deviation of 0.8. Conduct a hypothesis test to determine if the means are approximately the same or if the girls’ mean is higher than the boys’ mean.

Short Answer

Expert verified
The girls use significantly more ring tones than the boys.

Step by step solution

01

State the Hypotheses

First, we need to state our null and alternative hypotheses. The null hypothesis \( H_0 \) is that the mean number of ring tones for girls is equal to the mean number for boys. The alternative hypothesis \( H_1 \) is that the mean number of ring tones for girls is greater than the mean for boys. Thus, we have: - \( H_0: \mu_{\text{girls}} = \mu_{\text{boys}} \) - \( H_1: \mu_{\text{girls}} > \mu_{\text{boys}} \)
02

Choosing the Significance Level

Typically, a significance level \( \alpha \) is chosen to be 0.05, which indicates a 5% risk of concluding that a difference exists when there is no actual difference.
03

Calculate the Test Statistic

To calculate the test statistic, use the formula for the test statistic of two independent means: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] where \( \bar{x}_1 = 3.2 \), \( \bar{x}_2 = 1.7 \), \( s_1 = 1.5 \), \( s_2 = 0.8 \), \( n_1 = n_2 = 20 \). Substituting these values produces: \[ t = \frac{3.2 - 1.7}{\sqrt{\frac{1.5^2}{20} + \frac{0.8^2}{20}}} \] Calculating the values, the test statistic \( t \approx 4.021 \).
04

Determine the Critical Value

Determine the critical value from the t-distribution table with \( n_1 + n_2 - 2 = 38 \) degrees of freedom for a one-tailed test at \( \alpha = 0.05 \). The critical value of \( t\) is approximately 1.685.
05

Make the Decision

Compare the calculated test statistic \( t \approx 4.021 \) with the critical value \( t_{\alpha} = 1.685 \). Since \( t \approx 4.021 > t_{\alpha} = 1.685 \), we reject the null hypothesis.
06

Conclusion

Since the test statistic exceeds the critical value, there is sufficient evidence to support the claim that the mean number of ring tones for girls is higher than that for boys at a 5% significance level.

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Key Concepts

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

Independent Means
In the context of hypothesis testing, we often compare two groups to see if there's a significant difference between their means. Here, we are comparing the mean number of ring tones used by teenage girls and boys. These two groups are referred to as independent because the selection or characteristics of the girls do not influence or affect the selection or characteristics of the boys.

When we say 'independent means', we imply that each group is a distinct sample without overlap in membership. This is critical because it affects the calculations of variances and means used in the hypothesis test. For example, in our study, the girls with cellular phones and the boys with cellular phones are independent of each other. This means we can consider their sample statistics separately when calculating whether their means differ significantly.

Using independent samples is a common method for testing hypotheses when you want to understand the potential difference between two distinct groups.
Significance Level
The significance level, often denoted as \( \alpha \), is a key part of hypothesis testing. It represents the probability of rejecting the null hypothesis when it is actually true, known as a Type I error. In most studies, the significance level is set at 0.05, meaning there is a 5% risk of concluding that a difference exists when, in fact, there is none.

Choosing a significance level is about balancing the risks: a lower \( \alpha \) reduces the chance of a Type I error but increases the risk of a Type II error (failing to reject a false null hypothesis).

In our ringtone example, a significance level of 0.05 was chosen. This informs us that if our test statistic exceeds the critical value associated with this level, we have enough evidence to conclude that the mean number of ringtones for girls is statistically higher than that for boys.
Test Statistic Calculation
To understand whether the difference in means between two independent groups is statistically significant, we compute a test statistic. Here, we use the formula for the t-test for two independent means:
  • \( t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \)
Where \( \bar{x}_1 \) and \( \bar{x}_2 \) are the sample means, \( s_1 \) and \( s_2 \) are the sample standard deviations, and \( n_1 \) and \( n_2 \) are the sample sizes.

For the ringtone study, the means, standard deviations, and sample sizes were substituted into the formula, resulting in a calculated test statistic of approximately 4.021.

This test statistic measures how far the sample mean is from the population mean, in terms of standard errors. A higher absolute value of the test statistic suggests a greater statistical difference between groups.
Null and Alternative Hypotheses
The hypothesis testing process always begins with establishing the null and alternative hypotheses. The null hypothesis \( H_0 \) usually states that there is no effect or difference, which we wish to test against an alternative hypothesis \( H_1 \).

In our study of ring tones:
  • \( H_0: \mu_{\text{girls}} = \mu_{\text{boys}} \) (the means are equal)
  • \( H_1: \mu_{\text{girls}} > \mu_{\text{boys}} \) (the mean for girls is greater)
The null hypothesis reflects the common assumption that there's no difference between groups until proven otherwise. Conversely, the alternative hypothesis represents the claim that is being tested—in this case, that girls use significantly more ring tones than boys.

Setting up these hypotheses correctly is crucial as they provide the framework for statistical testing, guiding how we interpret test results and draw conclusions.

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Most popular questions from this chapter

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Use the following information to answer the next 12 exercises: The U.S. Center for Disease Control reports that the mean life expectancy was 47.6 years for whites born in 1900 and 33.0 years for nonwhites. Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124 whites, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 nonwhites, the mean life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test to see if the mean life spans in the county were the same for whites and nonwhites. Find the p-value.

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