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A cell phone company offers two plans to its subscribers. At the time new subscribers sign up, they are asked to provide some demographic information. The mean yearly income for a sample of 40 subscribers to Plan \(A\) is \(\$ 57,000\) with a standard deviation of \(\$ 9,200 .\) For a sample of 30 subscribers to Plan \(B\), the mean income is \(\$ 61,000\) with a standard deviation of \(\$ 7,100 .\) At the .05 significance level, is it reasonable to conclude the mean income of those selecting Plan \(\mathrm{B}\) is larger? What is the \(p\) -value?

Short Answer

Expert verified
Mean income for Plan B subscribers is not significantly larger. The p-value is 0.05.

Step by step solution

01

Define the Hypotheses

We need to identify the null and alternative hypotheses to determine if the mean income of Plan B subscribers is significantly larger than that of Plan A. - Null Hypothesis (H_0): \( \mu_B \leq \mu_A \)- Alternative Hypothesis (H_1): \( \mu_B > \mu_A \)
02

Identify the Test Statistic

Since we are comparing the means of two independent samples, the appropriate test statistic is the two-sample t-test for comparing means. The formula for the test statistic is:\[ t = \frac{\bar{x}_B - \bar{x}_A}{\sqrt{\frac{s_B^2}{n_B} + \frac{s_A^2}{n_A}}} \]where:- \( \bar{x}_A = 57000 \)- \( \bar{x}_B = 61000 \)- \( s_A = 9200 \)- \( s_B = 7100 \)- \( n_A = 40 \)- \( n_B = 30 \)
03

Calculate the Test Statistic

Substitute the given values into the formula:\[ t = \frac{61000 - 57000}{\sqrt{\frac{7100^2}{30} + \frac{9200^2}{40}}} \]Calculate the two terms in the denominator:1. \( \frac{7100^2}{30} \approx 1679000 / 30 = 55966.67 \)2. \( \frac{9200^2}{40} \approx 84640000 / 40 = 2116000 \)Denominator = \( \sqrt{55966.67 + 2116000} \approx 2394.14 \)Finally, calculate t:\[ t \approx \frac{4000}{2394.14} \approx 1.67 \]
04

Determine the Degrees of Freedom

For comparing two means with different variances, we use the formula for degrees of freedom in a two-sample t-test:\[ df = \frac{\left( \frac{s_A^2}{n_A} + \frac{s_B^2}{n_B} \right)^2}{\frac{\left( \frac{s_A^2}{n_A} \right)^2}{n_A - 1} + \frac{\left( \frac{s_B^2}{n_B} \right)^2}{n_B - 1}} \]Calculate the degrees of freedom, which approximately result in:\( df \approx 56 \) (rounding to the nearest whole number after calculation).
05

Find the Critical t-value

Using a t-table or calculator, find the critical t-value for \( \alpha = 0.05 \) and \( df \approx 56 \). This is a one-tailed test, so find for \( t_{0.05} \).The critical t-value is approximately 1.67.
06

Make the Decision

Compare the calculated t-value with the critical value: - Calculated t-value: 1.67 - Critical t-value: 1.67 Since the calculated t-value is equal to the critical value, we do not reject the null hypothesis.
07

Calculate the p-value

Using a t-distribution table or software, find the p-value for the calculated t-value (1.67) with approximately 56 degrees of freedom. The p-value is approximately 0.05.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a crucial statistical tool used to make decisions about data. It helps determine whether an observed effect is due to chance or significant enough to warrant further consideration. In this context, we are examining whether the mean income of those choosing Plan B is greater than those choosing Plan A.

To conduct a hypothesis test, you start with two statements:
  • **Null Hypothesis (H_0):** This suggests there is no significant difference or that any observed difference is due to sampling variation. For our problem, it is stated as \( \mu_B \leq \mu_A \)\.
  • **Alternative Hypothesis (H_1):** This indicates an actual effect or difference exists. In the example, it is \( \mu_B > \mu_A \)\.
After setting these hypotheses, the next steps involve calculating test statistics, which determine whether we should reject or fail to reject our null hypothesis.
Degrees of Freedom
Degrees of freedom (df) play a vital role in various statistical methods, including the t-test. They represent the number of independent elements that can vary within your sample data. In a two-sample t-test, they help adjust for sample size differences and are used to determine the appropriate distribution to refer to when calculating the p-value.The formula for calculating degrees of freedom in a two-sample t-test, especially when the variances of the samples are unequal, is:\[df = \frac{\left( \frac{s_A^2}{n_A} + \frac{s_B^2}{n_B} \right)^2}{\frac{\left( \frac{s_A^2}{n_A} \right)^2}{n_A - 1} + \frac{\left( \frac{s_B^2}{n_B} \right)^2}{n_B - 1}}\]

This expression accounts for the variability and size differences in the samples. In our classroom example, the degrees of freedom calculated approximately to 56, ensuring that we can accurately compare the statistics to the t-distribution for decision making.
P-Value
The p-value is a statistical metric that helps determine the significance of your hypothesis test results. It tells us the probability of observing a test statistic as extreme as, or more extreme than, the observed value, assuming the null hypothesis is true.

If the p-value is low (typically below the significance level like 0.05), it suggests that the observed data is inconsistent with the null hypothesis, leading us to reject the null hypothesis in favor of the alternative. In our two-sample t-test example, the calculated p-value was approximately 0.05.
  • If \( p \leq \alpha \), reject the null hypothesis.
  • If \( p > \alpha \), fail to reject the null hypothesis.
Here, since the p-value matches the significance level of 0.05, it falls right on the boundary of significance, making the decision a bit more nuanced. It suggests that, under these conditions, we do not have strong evidence to proclaim Plan B's mean income higher.
Significance Level
The significance level, often denoted by \( \alpha \), is a threshold set by the researcher to determine when an effect is statistically significant. It reflects the probability of rejecting the null hypothesis when it is actually true (a type I error). Commonly used significance levels include 0.01, 0.05, and 0.10.

In our example with the cell phone plans, we used a significance level of 0.05. This means we are willing to accept a 5% risk of incorrectly concluding that Plan B's mean income is higher than Plan A's. When we set \( \alpha = 0.05 \), it serves as a guideline to assess our p-value.
  • **\( p \leq \alpha \)**: Suggest significant results prompting rejection of \( H_0 \).
  • **\( p > \alpha \)**: Indicates insufficient evidence against \( H_0 \), so we fail to reject it.
This threshold guides researchers in making more informed decisions about their hypotheses.

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Most popular questions from this chapter

Clark Heter is an industrial engineer at Lyons Products. He would like to determine whether there are more units produced on the night shift than on the day shift. The mean number of units produced by a sample of 54 day-shift workers was \(345 .\) The mean number of units produced by a sample of 60 night- shift workers was 351 . Assume the population standard deviation of the number of units produced on the day shift is 21 and on the night shift is 28 . Using the .05 significance level, is the number of units produced on the night shift larger?

Fry Brothers Heating and Air Conditioning Inc. employs Larry Clark and George Murnen to make service calls to repair furnaces and air-conditioning units in homes. Tom Fry, the owner, would like to know whether there is a difference in the mean number of service calls they make per day. A random sample of 40 days last year showed that Larry Clark made an average of 4.77 calls per day. For a sample of 50 days, George Murnen made an average of 5.02 calls per day. Assume the population standard deviation is 1.05 calls per day for Larry Clark and 1.23 calls per day for George Murnen. At the .05 significance level, is there a difference in the mean number of calls per day between the two employees? What is the \(p\) -value?

Listed below are the 25 players on the opening-day roster of the 2016 New York Yankees Major League Baseball team, their salaries, and fielding positions. Sort the players into two groups, all pitchers (relief and starting) and position players (all others). Assume equal population standard deviations for the pitchers and the position players. Test the hypothesis that mean salaries of pitchers and position players are equal, using the . 01 significance level.

Gibbs Baby Food Company wishes to compare the weight gain of infants using its brand versus its competitor's. A sample of 40 babies using the Gibbs products revealed a mean weight gain of 7.6 pounds in the first three months after birth. For the Gibbs brand, the population standard deviation of the sample is 2.3 pounds. A sample of 55 babies using the competitor's brand revealed a mean increase in weight of 8.1 pounds. The population standard deviation is 2.9 pounds. At the .05 significance level, can we conclude that babies using the Gibbs brand gained less weight? Compute the \(p\) -value and interpret it.

Lester Hollar is vice president for human resources for a large manufacturing company. In recent years, he has noticed an increase in absenteeism that he thinks is related to the general health of the employees. Four years ago, in an attempt to improve the situation, he began a fitness program in which employees exercise during their lunch hour. To evaluate the program, he selected a random sample of eight participants and found the number of days each was absent in the six months before the exercise program began and in the six months following the exercise program. Below are the results. At the .05 significance level, can he conclude that the number of absences has declined? Estimate the \(p\) -value. $$ \begin{array}{|lcc|} \hline \text { Employee } & \text { Before } & \text { After } \\ \hline \text { Bauman } & 6 & 5 \\ \text { Briggs } & 6 & 2 \\ \text { Dottellis } & 7 & 1 \\ \text { Lee } & 7 & 3 \\ \text { Perralt } & 4 & 3 \\ \text { Rielly } & 3 & 6 \\ \text { Steinmetz } & 5 & 3 \\ \text { Stoltz } & 6 & 7 \\ \hline \end{array} $$

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