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Does the average length of time to earn a doctorate differ from one field to another? Independent random samples from large graduate schools gave the following averages for length of registered time (in years) from bachelor's degree to doctorate. Sample A was taken from the humanities field, and sample \(\mathrm{B}\) from the social sciences field. $$ \begin{array}{l|cccccccccccc} \hline \text { Field A } & 8.9 & 8.3 & 7.2 & 6.4 & 8.0 & 7.5 & 7.1 & 6.0 & 9.2 & 8.7 & 7.5 & \\ \hline \text { Field B } & 7.6 & 7.9 & 6.2 & 5.8 & 7.8 & 8.3 & 8.5 & 7.0 & 6.3 & 5.4 & 5.9 & 7.7 \\ \hline \end{array} $$ Use a \(1 \%\) level of significance to test the claim that there is no difference in the distributions of time to complete a doctorate for the two fields.

Short Answer

Expert verified
The average time to earn a doctorate does not differ significantly between the fields at the 1% level of significance.

Step by step solution

01

Define the Hypotheses

We need to conduct a hypothesis test to compare the means of the two independent samples. The null hypothesis (H_0) states that there is no difference in the average length of time to earn a doctorate between the two fields, i.e., \( \mu_A = \mu_B \). The alternative hypothesis (H_1) is that there is a difference, i.e., \( \mu_A eq \mu_B \).
02

Choose the Significance Level

The significance level is given as 1%, or \( \alpha = 0.01 \). This means we are allowing a 1% chance of rejecting the null hypothesis if it is actually true.
03

Calculate the Sample Means

Compute the average of each set of sample data. For Field A (Humanities), the mean is calculated as \( \bar{x}_A = \frac{8.9 + 8.3 + 7.2 + 6.4 + 8.0 + 7.5 + 7.1 + 6.0 + 9.2 + 8.7}{10} \). For Field B (Social Sciences), the mean is \( \bar{x}_B = \frac{7.6 + 7.9 + 6.2 + 5.8 + 7.8 + 8.3 + 8.5 + 7.0 + 6.3 + 5.4 + 5.9 + 7.7}{12} \).
04

Calculate the Sample Variances

Determine the variance for each field. For Field A, use \( s_A^2 = \frac{\sum (x_i - \bar{x}_A)^2}{n_A - 1} \) where \( n_A = 10 \). For Field B, use \( s_B^2 = \frac{\sum (x_i - \bar{x}_B)^2}{n_B - 1} \) where \( n_B = 12 \).
05

Conduct a t-test for Independent Samples

Use an independent sample t-test formula: \( t = \frac{\bar{x}_A - \bar{x}_B}{\sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}} \). This will allow us to find the calculated t statistic for the samples.
06

Determine the Degrees of Freedom

Calculate the degrees of freedom using the formula: \[ df = \frac{ \left( \frac{s_A^2}{n_A} + \frac{s_B^2}{n_B} \right)^2 }{ \frac{\left(\frac{s_A^2}{n_A}\right)^2}{n_A-1} + \frac{\left(\frac{s_B^2}{n_B}\right)^2}{n_B-1} } \].
07

Find the Critical t-value

Using a t-distribution table, find the critical t-value for a two-tailed test with \( \alpha = 0.01 \) and the calculated degrees of freedom from Step 6.
08

Compare t-value and Make a Decision

Compare the calculated t-value from Step 5 with the critical t-value from Step 7. If the calculated t-value exceeds the critical t-value, reject the null hypothesis. Otherwise, do not reject the null hypothesis.

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Key Concepts

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

t-test for independent samples
The t-test for independent samples is used to determine if there is a significant difference between the means of two unrelated groups. In this exercise, we are comparing the average time to doctorate completion between two different fields: humanities and social sciences. The goal of the test is to see whether the means of these groups are statistically different from each other.
The process begins by formulating two hypotheses: the null hypothesis (\(H_0\)) assumes there is no difference in means, while the alternative hypothesis (\(H_1\)) suggests that the means are different. This is like saying that the length of time to earn a doctorate is the same between the two fields versus being different.
This testing method involves calculating the t-statistic using the means and standard deviations of both samples. The formula for the t-statistic is:\[ t = \frac{\bar{x}_A - \bar{x}_B}{\sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}} \]where \(\bar{x}_A\) and \(\bar{x}_B\) are the means, \(s_A^2\) and \(s_B^2\) are the variances, and \(n_A\) and \(n_B\) are the sample sizes for groups A and B, respectively.
It's important to understand that the t-test for independent samples is useful when you have two separate groups, and you want to see if their averages are meaningfully different. You would not use this test for related groups, such as measurements taken at two different times on the same subjects.
significance level
The significance level is a crucial part of hypothesis testing, represented by the symbol \(\alpha\). It's the probability of rejecting the null hypothesis when it's true, which means it's the risk we are willing to take of making a type I error. A type I error occurs when we incorrectly conclude that there is a difference when there isn't one.
In this exercise, we have chosen a significance level of 1% or \(\alpha = 0.01\). This means we have a 1% risk of rejecting the null hypothesis by mistake. A lower significance level indicates a more stringent requirement for evidence against the null hypothesis. Therefore, choosing a significance level is a balance between being cautious and not missing out on detecting an actual effect.
This 1% threshold is particularly rigorous, indicating that very strong evidence is required to declare a significant difference in the average times for completing a doctorate between the two fields. It's all about setting a standard for the level of proof needed to accept the findings, essential for avoiding false positives.
degrees of freedom
Degrees of freedom are a concept used during hypothesis testing to determine the shape of the t-distribution, crucially affecting the critical value you'll use to make your decision. They reflect the number of independent values that can vary in the analysis without breaking any constraints.
In a t-test for independent samples, the degrees of freedom can be calculated using the formula:\[ df = \frac{ \left( \frac{s_A^2}{n_A} + \frac{s_B^2}{n_B} \right)^2 }{ \frac{\left(\frac{s_A^2}{n_A}\right)^2}{n_A-1} + \frac{\left(\frac{s_B^2}{n_B}\right)^2}{n_B-1} } \]This formula accounts for the variances and sample sizes of the two groups.
The degree of freedom affects the critical t-value that you obtain from the t-distribution table. With a higher degree of freedom, the t-distribution approaches more closely to the normal distribution.
  • A lower degree of freedom may reflect small sample sizes or a high number of estimated parameters, leading to a wider t-distribution.
  • A higher degree of freedom indicates more data points and usually results in a curve similar to the standard normal, leading to more precise critical values.
This concept is key in deciding whether your calculated t-value falls into the rejection region and thereby rejecting or failing to reject the null hypothesis.

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Most popular questions from this chapter

Consider the Spearman rank correlation coefficient \(r_{s}\) for data pairs \((x, y) .\) What is the monotone relationship, if any, between \(x\) and \(y\) implied by a value of (a) \(r_{s}=0 ?\) (b) \(r_{s}\) close to \(1 ?\) (c) \(r_{s}\) close to \(-1 ?\)

Are yields for organic farming different from conventional farming yields? Independent random samples from method A (organic farming) and method B (conventional farming) gave the following information about yield of lima beans (in tons/acre) (Reference: Agricultural Statistics, U.S. Department of Agriculture). $$ \begin{array}{l|llllllllllll} \hline \text { Method A } & 1.83 & 2.34 & 1.61 & 1.99 & 1.78 & 2.01 & 2.12 & 1.15 & 1.41 & 1.95 & 1.25 & \\ \hline \text { Method B } & 2.15 & 2.17 & 2.11 & 1.89 & 1.34 & 1.88 & 1.96 & 1.10 & 1.75 & 1.80 & 1.53 & 2.21 \\ \hline \end{array} $$ Use a \(5 \%\) level of significance to test the hypothesis that there is no difference between the yield distributions.

Is the crime rate in New York different from the crime rate in New Jersey? Independent random samples from region A (cities in New York) and region B (cities in New Jersey) gave the following information about violent crime rate (number of violent crimes per 100,000 population) (Reference: U.S. Department of Justice, Federal Bureau of Investigation). $$ \begin{array}{l|llllllllllll} \hline \text { Region A } & 554 & 517 & 492 & 561 & 577 & 621 & 512 & 580 & 543 & 605 & 531 & \\ \hline \text { Region B } & 475 & 419 & 505 & 575 & 395 & 433 & 521 & 388 & 375 & 411 & 586 & 415 \\ \hline \end{array} $$ Use a \(5 \%\) level of significance to test the claim that there is no difference in the crime rate distributions of the two states.

Is there a relation between police protection and fire protection? A random sample of large population areas gave the following information about the number of local police and the number of local firefighters (units in thousands). $$ \begin{array}{l|rrrrrrrrrrrrr} \hline \text { Area } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 \\\ \hline \text { Police } & 11.1 & 6.6 & 8.5 & 4.2 & 3.5 & 2.8 & 5.9 & 7.9 & 2.9 & 18.0 & 9.7 & 7.4 & 1.8 \\ \text { Firefighters } & 5.5 & 2.4 & 4.5 & 1.6 & 1.7 & 1.0 & 1.7 & 5.1 & 1.3 & 12.6 & 2.1 & 3.1 & 0.6 \\ \hline \end{array} $$ (i) Rank-order police using 1 as the largest data value. Also rank-order firefighters using 1 as the largest data value. Then construct a table of ranks to be used for a Spearman rank correlation test. (ii) Use a \(5 \%\) level of significance to test the claim that there is a monotone relationship (either way) between the ranks of number of police and number of firefighters.

Twenty-two fourth-grade children were randomly divided into two groups. Group A was taught spelling by a phonetic method. Group B was taught spelling by a memorization method. At the end of the fourth grade, all children were given a standard spelling exam. The scores are as follows. $$ \begin{array}{l|cccccccccccc} \hline \text { Group A } & 77 & 95 & 83 & 69 & 85 & 92 & 61 & 79 & 87 & 93 & 65 & 78 \\ \hline \text { Group B } & 62 & 90 & 70 & 81 & 63 & 75 & 80 & 72 & 82 & 94 & 65 & 79 \\ \hline \end{array} $$ Use a \(1 \%\) level of significance to test the claim that there is no difference in the test score distributions based on instruction method.

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