What test statistic can be used when the population standard deviation is unknown?

1. The two independent samples are simple random samples from two distinct populations.
2. For the two distinct populations:
   • if the sample sizes are small, the distributions are important (should be normal)
   • if the sample sizes are large, the distributions are not important (need not be normal)

The test comparing two independent population means with unknown and possibly unequal population standard deviations is called the Aspin-Welch t-test. The degrees of freedom formula was developed by Aspin-Welch.

The comparison of two population means is very common. A difference between the two samples depends on both the means and the standard deviations. Very different means can occur by chance if there is great variation among the individual samples. In order to account for the variation, we take the difference of the sample means, X¯ 1

– X¯2

, and divide by the standard error in order to standardize the difference. The result is a t-score test statistic.

Because we do not know the population standard deviations, we estimate them using the two sample standard deviations from our independent samples. For the hypothesis test, we calculate the estimated standard deviation, or standard error, of the difference in sample means, X¯1

– X¯2

.

The standard error is: (s1)2n1+(s2)2n2

The test statistic (t-score) is calculated as follows:

( x¯1–x¯2)–(μ1–μ2)(s1)2n1+ (s2)2n2

where:

• s1 and s2, the sample standard deviations, are estimates of σ1 and σ2, respectively.
• σ1 and σ1 are the unknown population standard deviations.
• x¯1

  and

  x¯2

  are the sample means. μ1 and μ2 are the population means.

The number of degrees of freedom (df) requires a somewhat complicated calculation. However, a computer or calculator calculates it easily. The df are not always a whole number. The test statistic calculated previously is approximated by the Student's t-distribution with df as follows:

Degrees of freedomdf= ((s1)2n1+(s2)2n2)2 (1n1–1)((s1)2n1)2+(1n 2–1)((s2)2n2)2

When both sample sizes n1 and n2 are five or larger, the Student's t approximation is very good. Notice that the sample variances (s1)2 and (s2)2 are not pooled. (If the question comes up, do not pool the variances.)

It is not necessary to compute this by hand. A calculator or computer easily computes it.

Independent groups

The average amount of time boys and girls aged seven to 11 spend playing sports each day is believed to be the same. A study is done and data are collected, resulting in the data in [link]. Each populations has a normal distribution.

| | Boys | 16 | 3.2 | 1.00 | {: valign=”top”}{: #uid888 summary=”This table presents the sample size in the second column, average hours a day in the third column, and the sample standard deviation in the fourth column. The first row is for girls and the second row is for boys.”}

Is there a difference in the mean amount of time boys and girls aged seven to 11 play sports each day? Test at the 5% level of significance.

The population standard deviations are not known. Let g be the subscript for girls and b be the subscript for boys. Then, μg is the population mean for girls and μb is the population mean for boys. This is a test of two independent groups, two population means.

Random variable: X¯g−X¯b

= difference in the sample mean amount of time girls and boys play sports each day. * * *

H0: μg = μb H0: μg – μb = 0 * * *

Ha: μg ≠ μb Ha: μg – μb ≠ 0 * * *

The words “the same” tell you H0 has an “=”. Since there are no other words to indicate Ha, assume it says “is different.” This is a two-tailed test.

Distribution for the test: Use tdf where df is calculated using the df formula for independent groups, two population means. Using a calculator, df is approximately 18.8462. Do not pool the variances.

Calculate the p-value using a Student’s t-distribution: p-value = 0.0054

Graph:

sg=0.866

sb=1

So, x¯g–x¯ b

= 2 – 3.2 = –1.2 * * *

Half the p-value is below –1.2 and half is above 1.2.

Make a decision: Since α > p-value, reject H0. This means you reject μg = μb. The means are different.

Press STAT. Arrow over to TESTS and press 4:2-SampTTest. Arrow over to Stats and press ENTER. Arrow down and enter 2 for the first sample mean, `

0.866

for Sx1, 9 for n1, 3.2 for the second sample mean, 1 for Sx2, and 16 for n2. Arrow down to μ1: and arrow to does not equal μ2. Press ENTER. Arrow down to Pooled: and No. Press ENTER. Arrow down to Calculate and press ENTER`. The p-value is p = 0.0054, the dfs are approximately 18.8462, and the test statistic is -3.14. Do the procedure again but instead of Calculate do Draw.

Conclusion: At the 5% level of significance, the sample data show there is sufficient evidence to conclude that the mean number of hours that girls and boys aged seven to 11 play sports per day is different (mean number of hours boys aged seven to 11 play sports per day is greater than the mean number of hours played by girls OR the mean number of hours girls aged seven to 11 play sports per day is greater than the mean number of hours played by boys).

Try It

Two samples are shown in [link]. Both have normal distributions. The means for the two populations are thought to be the same. Is there a difference in the means? Test at the 5% level of significance.

NOTE

When the sum of the sample sizes is larger than 30 (n1 + n2 > 30) you can use the normal distribution to approximate the Student's t.

A study is done by a community group in two neighboring colleges to determine which one graduates students with more math classes. College A samples 11 graduates. Their average is four math classes with a standard deviation of 1.5 math classes. College B samples nine graduates. Their average is 3.5 math classes with a standard deviation of one math class. The community group believes that a student who graduates from college A has taken more math classes, on the average. Both populations have a normal distribution. Test at a 1% significance level. Answer the following questions.* * *

a. Is this a test of two means or two proportions?

a. two means* * *

b. Are the populations standard deviations known or unknown?

b. unknown* * *

c. Which distribution do you use to perform the test?

c. Student’s *t** * *

d. What is the random variable?

e. What are the null and alternate hypotheses? Write the null and alternate hypotheses in words and in symbols.

e.

• Ho:μA≤μB
• Ha :μA>μB

f. Is this test right-, left-, or two-tailed?

f.

right* * *

g. What is the p-value?

g. 0.1928* * *

h. Do you reject or not reject the null hypothesis?

h. Do not reject.* * *

i. Conclusion:

i. At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a student who graduates from college A has taken more math classes, on the average, than a student who graduates from college B.

Try It

A study is done to determine if Company A retains its workers longer than Company B. Company A samples 15 workers, and their average time with the company is five years with a standard deviation of 1.2. Company B samples 20 workers, and their average time with the company is 4.5 years with a standard deviation of 0.8. The populations are normally distributed.

1. Are the population standard deviations known?
2. Conduct an appropriate hypothesis test. At the 5% significance level, what is your conclusion?

A professor at a large community college wanted to determine whether there is a difference in the means of final exam scores between students who took his statistics course online and the students who took his face-to-face statistics class. He believed that the mean of the final exam scores for the online class would be lower than that of the face-to-face class. Was the professor correct? The randomly selected 30 final exam scores from each group are listed in [link] and [link].

Is the mean of the Final Exam scores of the online class lower than the mean of the Final Exam scores of the face-to-face class? Test at a 5% significance level. Answer the following questions:

1. Is this a test of two means or two proportions?
2. Are the population standard deviations known or unknown?
3. Which distribution do you use to perform the test?
4. What is the random variable?
5. What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols.
6. Is this test right, left, or two tailed?
7. What is the p-value?
8. Do you reject or not reject the null hypothesis?
9. At the ___ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______.

(See the conclusion in [link], and write yours in a similar fashion)

First put the data for each group into two lists (such as L1 and L2). Press STAT. Arrow over to TESTS and press 4:2SampTTest. Make sure Data is highlighted and press ENTER. Arrow down and enter L1 for the first list and L2 for the second list. Arrow down to μ1: and arrow to ≠ μ2 (does not equal). Press ENTER. Arrow down to Pooled: No. Press ENTER. Arrow down to Calculate and press ENTER.

Note

Be careful not to mix up the information for Group 1 and Group 2!

1. two means
2. unknown
3. Student’s t
4. X¯1–X¯2
5. 1. H0: μ1 = μ2 Null hypothesis: the means of the final exam scores are equal for the online and face-to-face statistics classes.
   2. Ha: μ1 < μ2 Alternative hypothesis: the mean of the final exam scores of the online class is less than the mean of the final exam scores of the face-to-face class.
6. left-tailed

7. p-value = 0.0011

   
   {:}

8. Reject the null hypothesis
9. The professor was correct. The evidence shows that the mean of the final exam scores for the online class is lower than that of the face-to-face class.

   ---

   At the 5% level of significance, from the sample data, there is (is/is not) sufficient evidence to conclude that the mean of the final exam scores for the online class is less than the mean of final exam scores of the face-to-face class.

Cohen's Standards for Small, Medium, and Large Effect SizesCohen's d is a measure of effect size based on the differences between two means. Cohen’s d, named for United States statistician Jacob Cohen, measures the relative strength of the differences between the means of two populations based on sample data. The calculated value of effect size is then compared to Cohen’s standards of small, medium, and large effect sizes.

Cohen's d is the measure of the difference between two means divided by the pooled standard deviation: d=x¯1–x¯2spooled

where spooled=(n1–1)s12+(n2–1) s22n1+n2–2

Calculate Cohen’s d for [link]. Is the size of the effect small, medium, or large? Explain what the size of the effect means for this problem.

μ1 = 4 s1 = 1.5 n1 = 11 * * *

μ2 = 3.5 s2 = 1 n2 = 9 * * *

d = 0.384 * * *

The effect is small because 0.384 is between Cohen’s value of 0.2 for small effect size and 0.5 for medium effect size. The size of the differences of the means for the two colleges is small indicating that there is not a significant difference between them.

Calculate Cohen’s d for [link]. Is the size of the effect small, medium or large? Explain what the size of the effect means for this problem.

d = 0.834; Large, because 0.834 is greater than Cohen’s 0.8 for a large effect size. The size of the differences between the means of the Final Exam scores of online students and students in a face-to-face class is large indicating a significant difference.

Try It

Weighted alpha is a measure of risk-adjusted performance of stocks over a period of a year. A high positive weighted alpha signifies a stock whose price has risen while a small positive weighted alpha indicates an unchanged stock price during the time period. Weighted alpha is used to identify companies with strong upward or downward trends. The weighted alpha for the top 30 stocks of banks in the northeast and in the west as identified by Nasdaq on May 24, 2013 are listed in [link] and [link], respectively.

Is there a difference in the weighted alpha of the top 30 stocks of banks in the northeast and in the west? Test at a 5% significance level. Answer the following questions:

1. Is this a test of two means or two proportions?
2. Are the population standard deviations known or unknown?
3. Which distribution do you use to perform the test?
4. What is the random variable?
5. What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols.
6. Is this test right, left, or two tailed?
7. What is the p-value?
8. Do you reject or not reject the null hypothesis?
9. At the ___ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______.
10. Calculate Cohen’s d and interpret it.

References

Data from Graduating Engineer + Computer Careers. Available online at http://www.graduatingengineer.com

Data from Microsoft Bookshelf.

Data from the United States Senate website, available online at www.Senate.gov (accessed June 17, 2013).

“List of current United States Senators by Age.” Wikipedia. Available online at http://en.wikipedia.org/wiki/List\_of\_current\_United\_States\_Senators\_by\_age (accessed June 17, 2013).

“Sectoring by Industry Groups.” Nasdaq. Available online at http://www.nasdaq.com/markets/barchart-sectors.aspx?page=sectors&base=industry (accessed June 17, 2013).

“Strip Clubs: Where Prostitution and Trafficking Happen.” Prostitution Research and Education, 2013. Available online at www.prostitutionresearch.com/ProsViolPosttrauStress.html (accessed June 17, 2013).

“World Series History.” Baseball-Almanac, 2013. Available online at http://www.baseball-almanac.com/ws/wsmenu.shtml (accessed June 17, 2013).

Chapter Review

Two population means from independent samples where the population standard deviations are not known

• Random Variable: X¯1−X¯2

  = the difference of the sampling means

• Distribution: Student's t-distribution with degrees of freedom (variances not pooled)

Formula Review

Standard error: SE = (s1)2n1+(s2)2n2

Test statistic (t-score): t = (x¯1−x¯2)−( μ1−μ2)(s1)2n1+(s2)2n 2

Degrees of freedom:* * *

where:

s1 and s2 are the sample standard deviations, and n1 and n2 are the sample sizes.

and x¯2

are the sample means.

Cohen’s d is the measure of effect size:

where spooled=(n1−1)s12+(n2−1)s22n1+n2−2

Homework

DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.

NOTE

If you are using a Student's t-distribution for a homework problem in what follows, including for paired data, you may assume that the underlying population is normally distributed. (When using these tests in a real situation, you must first prove that assumption, however.)

The mean number of English courses taken in a two–year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from 29 males and 16 females. The males took an average of three English courses with a standard deviation of 0.8. The females took an average of four English courses with a standard deviation of 1.0. Are the means statistically the same?

A student at a four-year college claims that mean enrollment at four–year colleges is higher than at two–year colleges in the United States. Two surveys are conducted. Of the 35 two–year colleges surveyed, the mean enrollment was 5,068 with a standard deviation of 4,777. Of the 35 four-year colleges surveyed, the mean enrollment was 5,466 with a standard deviation of 8,191.

Subscripts: 1: two-year colleges; 2: four-year colleges

1. H0: μ1 ≥ μ2
2. Ha: μ1 < μ2
3. X¯1–X¯2

   is the difference between the mean enrollments of the two-year colleges and the four-year colleges.

4. Student’s-t
5. test statistic: -0.2480
6. p-value: 0.4019
7. Check student’s solution.
8. 1. Alpha: 0.05
   2. Decision: Do not reject
   3. Reason for Decision: p-value > alpha
   4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean enrollment at four-year colleges is higher than at two-year colleges.

At Rachel’s 11th birthday party, eight girls were timed to see how long (in seconds) they could hold their breath in a relaxed position. After a two-minute rest, they timed themselves while jumping. The girls thought that the mean difference between their jumping and relaxed times would be zero. Test their hypothesis.

Mean entry-level salaries for college graduates with mechanical engineering degrees and electrical engineering degrees are believed to be approximately the same. A recruiting office thinks that the mean mechanical engineering salary is actually lower than the mean electrical engineering salary. The recruiting office randomly surveys 50 entry level mechanical engineers and 60 entry level electrical engineers. Their mean salaries were $46,100 and $46,700, respectively. Their standard deviations were $3,450 and $4,210, respectively. Conduct a hypothesis test to determine if you agree that the mean entry-level mechanical engineering salary is lower than the mean entry-level electrical engineering salary.

Subscripts: 1: mechanical engineering; 2: electrical engineering

1. H0: µ1 ≥ µ2
2. Ha: µ1 < µ2
3. X¯1−X¯2

   is the difference between the mean entry level salaries of mechanical engineers and electrical engineers.

4. t108
5. test statistic: t = –0.82
6. p-value: 0.2061
7. Check student’s solution.
8. 1. Alpha: 0.05
   2. Decision: Do not reject the null hypothesis.
   3. Reason for Decision: p-value > alpha
   4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean entry-level salaries of mechanical engineers is lower than that of electrical engineers.

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.

Use the information from [link] to answer the next four exercises.

Using the data from Lap 1 only, conduct a hypothesis test to determine if the mean time for completing a lap in races is the same as it is in practices.

1. H0: µ1 = µ2
2. Ha: µ1 ≠ µ2
3. X¯1−X¯2

   is the difference between the mean times for completing a lap in races and in practices.

4. t20.32
5. test statistic: –4.70
6. p-value: 0.0001
7. Check student’s solution.
8. 1. Alpha: 0.05
   2. Decision: Reject the null hypothesis.
   3. Reason for Decision: p-value < alpha
   4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean time for completing a lap in races is different from that in practices.

Repeat the test in Exercise 10.83, but this time combine the data from Laps 1 and 5.

1. H0: µ1 = µ2
2. Ha: µ1 ≠ µ2
3. is the difference between the mean times for completing a lap in races and in practices.
4. t40.94
5. test statistic: –5.08
6. p-value: zero
7. Check student’s solution.
8. 1. Alpha: 0.05
   2. Decision: Reject the null hypothesis.
   3. Reason for Decision: p-value < alpha
   4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean time for completing a lap in races is different from that in practices.

In two to three complete sentences, explain in detail how you might use Terri Vogel’s data to answer the following question. “Does Terri Vogel drive faster in races than she does in practices?”

Use the following information to answer the next two exercises. The Eastern and Western Major League Soccer conferences have a new Reserve Division that allows new players to develop their skills. Data for a randomly picked date showed the following annual goals.

Conduct a hypothesis test to answer the next two exercises.

The exact distribution for the hypothesis test is:

1. the normal distribution
2. the Student’s t-distribution
3. the uniform distribution
4. the exponential distribution

If the level of significance is 0.05, the conclusion is:

1. There is sufficient evidence to conclude that the W Division teams score fewer goals, on average, than the E teams
2. There is insufficient evidence to conclude that the W Division teams score more goals, on average, than the E teams.
3. There is insufficient evidence to conclude that the W teams score fewer goals, on average, than the E teams score.
4. Unable to determine

c

Suppose a statistics instructor believes that there is no significant difference between the mean class scores of statistics day students on Exam 2 and statistics night students on Exam 2. She takes random samples from each of the populations. The mean and standard deviation for 35 statistics day students were 75.86 and 16.91. The mean and standard deviation for 37 statistics night students were 75.41 and 19.73. The “day” subscript refers to the statistics day students. The “night” subscript refers to the statistics night students. A concluding statement is:

1. There is sufficient evidence to conclude that statistics night students’ mean on Exam 2 is better than the statistics day students’ mean on Exam 2.
2. There is insufficient evidence to conclude that the statistics day students’ mean on Exam 2 is better than the statistics night students’ mean on Exam 2.
3. There is insufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2.
4. There is sufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2.

Researchers interviewed street prostitutes in Canada and the United States. The mean age of the 100 Canadian prostitutes upon entering prostitution was 18 with a standard deviation of six. The mean age of the 130 United States prostitutes upon entering prostitution was 20 with a standard deviation of eight. Is the mean age of entering prostitution in Canada lower than the mean age in the United States? Test at a 1% significance level.

Test: two independent sample means, population standard deviations unknown.

Random variable: X¯1−X¯2

Distribution: H0: μ1 = μ2 Ha: μ1 < μ2 The mean age of entering prostitution in Canada is lower than the mean age in the United States.

Graph: left-tailed

p-value : 0.0151

Decision: Do not reject H0.

Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of entering prostitution in Canada is lower than the mean age in the United States.

A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. Of interest is whether the liquid diet yields a higher mean weight loss than the powder diet. The powder diet group had a mean weight loss of 42 pounds with a standard deviation of 12 pounds. The liquid diet group had a mean weight loss of 45 pounds with a standard deviation of 14 pounds.

Suppose a statistics instructor believes that there is no significant difference between the mean class scores of statistics day students on Exam 2 and statistics night students on Exam 2. She takes random samples from each of the populations. The mean and standard deviation for 35 statistics day students were 75.86 and 16.91, respectively. The mean and standard deviation for 37 statistics night students were 75.41 and 19.73. The “day” subscript refers to the statistics day students. The “night” subscript refers to the statistics night students. An appropriate alternative hypothesis for the hypothesis test is:

1. μday > μnight
2. μday < μnight
3. μday = μnight
4. μday ≠ μnight

d

Glossary

• The domain of the random variable (RV) is not necessarily a numerical set; the domain may be expressed in words; for example, if X = hair color, then the domain is {black, blond, gray, green, orange}.
• We can tell what specific value x of the random variable X takes only after performing the experiment.

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What test statistic formula will you use if the population standard deviation is unknown?

Which test to use when standard deviation is not given?

Can z

When the population standard deviation is unknown we use?