Suppose we wish to compare the means of two distinct populations. Figure 9.1 "Independent Sampling from Two Populations" illustrates the conceptual framework of our investigation in this and the next section. Each population has a mean and a standard deviation. We arbitrarily label one population as Population 1 and the other as Population 2, and subscript the parameters with the numbers 1 and 2 to tell them apart. We draw a random sample from Population 1 and label the sample statistics it yields with the subscript 1. Without reference to the first sample we draw a sample from Population 2 and label its sample statistics with the subscript 2.
Figure 9.1 Independent Sampling from Two Populations
Samples from two distinct populations are independent if each one is drawn without reference to the other, and has no connection with the other.
Our goal is to use the information in the samples to estimate the difference ${\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}$ in the means of the two populations and to make statistically valid inferences about it.
Since the mean ${\stackrel{-}{x}}_{1}$ of the sample drawn from Population 1 is a good estimator of ${\mathit{\mu}}_{1}$ and the mean ${\stackrel{-}{x}}_{2}$ of the sample drawn from Population 2 is a good estimator of ${\mathit{\mu}}_{2}$, a reasonable point estimate of the difference ${\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}$ is ${\stackrel{-}{x}}_{1}-{\stackrel{-}{x}}_{2}.$ In order to widen this point estimate into a confidence interval, we first suppose that both samples are large, that is, that both ${n}_{1}\ge 30$ and ${n}_{2}\ge 30.$ If so, then the following formula for a confidence interval for ${\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}$ is valid. The symbols ${s}_{1}^{2}$ and ${s}_{2}^{2}$ denote the squares of s_{1} and s_{2}. (In the relatively rare case that both population standard deviations ${\mathit{\sigma}}_{1}$ and ${\mathit{\sigma}}_{2}$ are known they would be used instead of the sample standard deviations.)
The samples must be independent, and each sample must be large: ${n}_{1}\ge 30$ and ${n}_{2}\ge 30.$
To compare customer satisfaction levels of two competing cable television companies, 174 customers of Company 1 and 355 customers of Company 2 were randomly selected and were asked to rate their cable companies on a five-point scale, with 1 being least satisfied and 5 most satisfied. The survey results are summarized in the following table:
Company 1 | Company 2 |
---|---|
${n}_{1}=174$ | ${n}_{2}=355$ |
${\stackrel{-}{x}}_{1}=3.51$ | ${\stackrel{-}{x}}_{2}=3.24$ |
${s}_{1}=0.51$ | ${s}_{2}=0.52$ |
Construct a point estimate and a 99% confidence interval for ${\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}$, the difference in average satisfaction levels of customers of the two companies as measured on this five-point scale.
Solution:
The point estimate of ${\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}$ is
$${\stackrel{-}{x}}_{1}-{\stackrel{-}{x}}_{2}=3.51-3.24=0.27.$$In words, we estimate that the average customer satisfaction level for Company 1 is 0.27 points higher on this five-point scale than it is for Company 2.
To apply the formula for the confidence interval, proceed exactly as was done in Chapter 7 "Estimation". The 99% confidence level means that $\alpha =1-0.99=0.01$ so that ${z}_{\alpha \u22152}={z}_{0.005}.$ From Figure 12.3 "Critical Values of " we read directly that ${z}_{0.005}=2.576.$ Thus
$$\left({\stackrel{-}{x}}_{1}-{\stackrel{-}{x}}_{2}\right)\pm {z}_{\alpha \u22152}\sqrt{\frac{{s}_{1}^{2}}{{n}_{1}}+\frac{{s}_{2}^{2}}{{n}_{2}}}=0.27\pm 2.576\sqrt{\frac{0.5{1}^{2}}{174}+\frac{0.5{2}^{2}}{355}}=0.27\pm 0.12$$We are 99% confident that the difference in the population means lies in the interval $\left[0.\mathrm{15,0}.39\right]$, in the sense that in repeated sampling 99% of all intervals constructed from the sample data in this manner will contain ${\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}.$ In the context of the problem we say we are 99% confident that the average level of customer satisfaction for Company 1 is between 0.15 and 0.39 points higher, on this five-point scale, than that for Company 2.
Hypotheses concerning the relative sizes of the means of two populations are tested using the same critical value and p-value procedures that were used in the case of a single population. All that is needed is to know how to express the null and alternative hypotheses and to know the formula for the standardized test statistic and the distribution that it follows.
The null and alternative hypotheses will always be expressed in terms of the difference of the two population means. Thus the null hypothesis will always be written
$${H}_{0}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}={D}_{0}$$where D_{0} is a number that is deduced from the statement of the situation. As was the case with a single population the alternative hypothesis can take one of the three forms, with the same terminology:
Form of ${H}_{a}$ | Terminology |
---|---|
${H}_{a}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}<{D}_{0}$ | Left-tailed |
${H}_{a}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}>{D}_{0}$ | Right-tailed |
${H}_{a}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}\ne {D}_{0}$ | Two-tailed |
As long as the samples are independent and both are large the following formula for the standardized test statistic is valid, and it has the standard normal distribution. (In the relatively rare case that both population standard deviations ${\mathit{\sigma}}_{1}$ and ${\mathit{\sigma}}_{2}$ are known they would be used instead of the sample standard deviations.)
The test statistic has the standard normal distribution.
The samples must be independent, and each sample must be large: ${n}_{1}\ge 30$ and ${n}_{2}\ge 30.$
Refer to Note 9.4 "Example 1" concerning the mean satisfaction levels of customers of two competing cable television companies. Test at the 1% level of significance whether the data provide sufficient evidence to conclude that Company 1 has a higher mean satisfaction rating than does Company 2. Use the critical value approach.
Solution:
Step 1. If the mean satisfaction levels ${\mathit{\mu}}_{1}$ and ${\mathit{\mu}}_{2}$ are the same then ${\mathit{\mu}}_{1}={\mathit{\mu}}_{2}$, but we always express the null hypothesis in terms of the difference between ${\mathit{\mu}}_{1}$ and ${\mathit{\mu}}_{2}$, hence H_{0} is ${\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}=0.$ To say that the mean customer satisfaction for Company 1 is higher than that for Company 2 means that ${\mathit{\mu}}_{1}>{\mathit{\mu}}_{2}$, which in terms of their difference is ${\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}>0.$ The test is therefore
$$\begin{array}{ccc}\hfill {H}_{0}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}& =& 0\\ \hfill \text{\hspace{0.17em}vs.\hspace{0.17em}}\text{\hspace{0.17em}}{H}_{a}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}& >& 0& @\text{\hspace{0.17em}}\alpha =0.01\end{array}$$Step 2. Since the samples are independent and both are large the test statistic is
$$Z=\frac{\left({\stackrel{-}{x}}_{1}-{\stackrel{-}{x}}_{2}\right)-{D}_{0}}{\sqrt{\frac{{s}_{1}^{2}}{{n}_{1}}+\frac{{s}_{2}^{2}}{{n}_{2}}}}$$Step 3. Inserting the data into the formula for the test statistic gives
$$Z=\frac{\left({\stackrel{-}{x}}_{1}-{\stackrel{-}{x}}_{2}\right)-{D}_{0}}{\sqrt{\frac{{s}_{1}^{2}}{{n}_{1}}+\frac{{s}_{2}^{2}}{{n}_{2}}}}=\frac{\left(3.51-3.24\right)\text{\u2212}0}{\sqrt{\frac{0.5{1}^{2}}{174}+\frac{0.5{2}^{2}}{355}}}=5.684$$Step 4. Since the symbol in H_{a} is “>” this is a right-tailed test, so there is a single critical value, ${z}_{\alpha}={z}_{0.01}$, which from the last line in Figure 12.3 "Critical Values of " we read off as 2.326. The rejection region is $\left[2.326,\infty \right).$
Figure 9.2 Rejection Region and Test Statistic for Note 9.6 "Example 2"
Step 5. As shown in Figure 9.2 "Rejection Region and Test Statistic for " the test statistic falls in the rejection region. The decision is to reject H_{0}. In the context of the problem our conclusion is:
The data provide sufficient evidence, at the 1% level of significance, to conclude that the mean customer satisfaction for Company 1 is higher than that for Company 2.
Perform the test of Note 9.6 "Example 2" using the p-value approach.
Solution:
The first three steps are identical to those in Note 9.6 "Example 2".
Figure 9.3 P-Value for Note 9.7 "Example 3"
Step 5. Since 0.0000 < 0.01, $p\text{\hspace{0.17em}-value}<\alpha $ so the decision is to reject the null hypothesis:
The data provide sufficient evidence, at the 1% level of significance, to conclude that the mean customer satisfaction for Company 1 is higher than that for Company 2.
Construct the confidence interval for ${\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}$ for the level of confidence and the data from independent samples given.
90% confidence,
${n}_{1}=45$, ${\stackrel{-}{x}}_{1}=27$, ${s}_{1}=2$
${n}_{2}=60$, ${\stackrel{-}{x}}_{2}=22$, ${s}_{2}=3$
99% confidence,
${n}_{1}=30$, ${\stackrel{-}{x}}_{1}=\text{\u2212}112$, ${s}_{1}=9$
${n}_{2}=40$, ${\stackrel{-}{x}}_{2}=\text{\u2212}98$, ${s}_{2}=4$
Construct the confidence interval for ${\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}$ for the level of confidence and the data from independent samples given.
95% confidence,
${n}_{1}=110$, ${\stackrel{-}{x}}_{1}=77$, ${s}_{1}=15$
${n}_{2}=85$, ${\stackrel{-}{x}}_{2}=79$, ${s}_{2}=21$
90% confidence,
${n}_{1}=65$, ${\stackrel{-}{x}}_{1}=\text{\u2212}83$, ${s}_{1}=12$
${n}_{2}=65$, ${\stackrel{-}{x}}_{2}=\text{\u2212}74$, ${s}_{2}=8$
Construct the confidence interval for ${\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}$ for the level of confidence and the data from independent samples given.
99.5% confidence,
${n}_{1}=130$, ${\stackrel{-}{x}}_{1}=27.2$, ${s}_{1}=2.5$
${n}_{2}=155$, ${\stackrel{-}{x}}_{2}=38.8$, ${s}_{2}=4.6$
95% confidence,
${n}_{1}=68$, ${\stackrel{-}{x}}_{1}=215.5$, ${s}_{1}=12.3$
${n}_{2}=84$, ${\stackrel{-}{x}}_{2}=287.8$, ${s}_{2}=14.1$
Construct the confidence interval for ${\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}$ for the level of confidence and the data from independent samples given.
99.9% confidence,
${n}_{1}=275$, ${\stackrel{-}{x}}_{1}=70.2$, ${s}_{1}=1.5$
${n}_{2}=325$, ${\stackrel{-}{x}}_{2}=63.4$, ${s}_{2}=1.1$
90% confidence,
${n}_{1}=120$, ${\stackrel{-}{x}}_{1}=35.5$, ${s}_{1}=0.75$
${n}_{2}=146$, ${\stackrel{-}{x}}_{2}=29.6$, ${s}_{2}=0.80$
Perform the test of hypotheses indicated, using the data from independent samples given. Use the critical value approach. Compute the p-value of the test as well.
Test ${H}_{0}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}=3$ vs. ${H}_{a}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}\ne 3$ @ $\alpha =0.05$,
${n}_{1}=35$, ${\stackrel{-}{x}}_{1}=25$, ${s}_{1}=1$
${n}_{2}=45$, ${\stackrel{-}{x}}_{2}=19$, ${s}_{2}=2$
Test ${H}_{0}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}=\text{\u2212}25$ vs. ${H}_{a}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}<\text{\u2212}25$ @ $\alpha =0.10$,
${n}_{1}=85$, ${\stackrel{-}{x}}_{1}=188$, ${s}_{1}=15$
${n}_{2}=62$, ${\stackrel{-}{x}}_{2}=215$, ${s}_{2}=19$
Perform the test of hypotheses indicated, using the data from independent samples given. Use the critical value approach. Compute the p-value of the test as well.
Test ${H}_{0}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}=45$ vs. ${H}_{a}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}>45$ @ $\alpha =0.001$,
${n}_{1}=200$, ${\stackrel{-}{x}}_{1}=1312$, ${s}_{1}=35$
${n}_{2}=225$, ${\stackrel{-}{x}}_{2}=1256$, ${s}_{2}=28$
Test ${H}_{0}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}=\text{\u2212}12$ vs. ${H}_{a}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}\ne \text{\u2212}12$ @ $\alpha =0.10$,
${n}_{1}=35$, ${\stackrel{-}{x}}_{1}=121$, ${s}_{1}=6$
${n}_{2}=40$, ${\stackrel{-}{x}}_{2}=135$, ${s}_{2}=7$
Perform the test of hypotheses indicated, using the data from independent samples given. Use the critical value approach. Compute the p-value of the test as well.
Test ${H}_{0}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}=0$ vs. ${H}_{a}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}\ne 0$ @ $\alpha =0.01$,
${n}_{1}=125$, ${\stackrel{-}{x}}_{1}=\text{\u2212}46$, ${s}_{1}=10$
${n}_{2}=90$, ${\stackrel{-}{x}}_{2}=\text{\u2212}50$, ${s}_{2}=13$
Test ${H}_{0}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}=20$ vs. ${H}_{a}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}>20$ @ $\alpha =0.05$,
${n}_{1}=40$, ${\stackrel{-}{x}}_{1}=142$, ${s}_{1}=11$
${n}_{2}=40$, ${\stackrel{-}{x}}_{2}=118$, ${s}_{2}=10$
Perform the test of hypotheses indicated, using the data from independent samples given. Use the critical value approach. Compute the p-value of the test as well.
Test ${H}_{0}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}=13$ vs. ${H}_{a}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}<13$ @ $\alpha =0.01$,
${n}_{1}=35$, ${\stackrel{-}{x}}_{1}=100$, ${s}_{1}=2$
${n}_{2}=35$, ${\stackrel{-}{x}}_{2}=88$, ${s}_{2}=2$
Test ${H}_{0}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}=\text{\u2212}10$ vs. ${H}_{a}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}\ne \text{\u2212}10$ @ $\alpha =0.10$,
${n}_{1}=146$, ${\stackrel{-}{x}}_{1}=62$, ${s}_{1}=4$
${n}_{2}=120$, ${\stackrel{-}{x}}_{2}=73$, ${s}_{2}=7$
Perform the test of hypotheses indicated, using the data from independent samples given. Use the p-value approach.
Test ${H}_{0}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}=57$ vs. ${H}_{a}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}<57$ @ $\alpha =0.10$,
${n}_{1}=117$, ${\stackrel{-}{x}}_{1}=1309$, ${s}_{1}=42$
${n}_{2}=133$, ${\stackrel{-}{x}}_{2}=1258$, ${s}_{2}=37$
Test ${H}_{0}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}=\text{\u2212}1.5$ vs. ${H}_{a}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}\ne \text{\u2212}1.5$ @ $\alpha =0.20$,
${n}_{1}=65$, ${\stackrel{-}{x}}_{1}=16.9$, ${s}_{1}=1.3$
${n}_{2}=57$, ${\stackrel{-}{x}}_{2}=18.6$, ${s}_{2}=1.1$
Perform the test of hypotheses indicated, using the data from independent samples given. Use the p-value approach.
Test ${H}_{0}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}=\text{\u2212}10.5$ vs. ${H}_{a}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}>\text{\u2212}10.5$ @ $\alpha =0.01$,
${n}_{1}=64$, ${\stackrel{-}{x}}_{1}=85.6$, ${s}_{1}=2.4$
${n}_{2}=50$, ${\stackrel{-}{x}}_{2}=95.3$, ${s}_{2}=3.1$
Test ${H}_{0}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}=110$ vs. ${H}_{a}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}\ne 110$ @ $\alpha =0.02$,
${n}_{1}=176$, ${\stackrel{-}{x}}_{1}=1918$, ${s}_{1}=68$
${n}_{2}=241$, ${\stackrel{-}{x}}_{2}=1782$, ${s}_{2}=146$
Perform the test of hypotheses indicated, using the data from independent samples given. Use the p-value approach.
Test ${H}_{0}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}=50$ vs. ${H}_{a}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}>50$ @ $\alpha =0.005$,
${n}_{1}=72$, ${\stackrel{-}{x}}_{1}=272$, ${s}_{1}=26$
${n}_{2}=103$, ${\stackrel{-}{x}}_{2}=213$, ${s}_{2}=14$
Test ${H}_{0}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}=7.5$ vs. ${H}_{a}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}\ne 7.5$ @ $\alpha =0.10$,
${n}_{1}=52$, ${\stackrel{-}{x}}_{1}=94.3$, ${s}_{1}=2.6$
${n}_{2}=38$, ${\stackrel{-}{x}}_{2}=88.6$, ${s}_{2}=8.0$
Perform the test of hypotheses indicated, using the data from independent samples given. Use the p-value approach.
Test ${H}_{0}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}=23$ vs. ${H}_{a}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}<23$ @ $\alpha =0.20$,
${n}_{1}=314$, ${\stackrel{-}{x}}_{1}=198$, ${s}_{1}=12.2$
${n}_{2}=220$, ${\stackrel{-}{x}}_{2}=176$, ${s}_{2}=11.5$
Test ${H}_{0}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}=4.4$ vs. ${H}_{a}:{\mathit{\mu}}_{1}-{\mathit{\mu}}_{2}\ne 4.4$ @ $\alpha =0.05$,
${n}_{1}=32$, ${\stackrel{-}{x}}_{1}=40.3$, ${s}_{1}=0.5$
${n}_{2}=30$, ${\stackrel{-}{x}}_{2}=35.5$, ${s}_{2}=0.7$
In order to investigate the relationship between mean job tenure in years among workers who have a bachelor’s degree or higher and those who do not, random samples of each type of worker were taken, with the following results.
n | $\stackrel{-}{x}$ | s | |
---|---|---|---|
Bachelor’s degree or higher | 155 | 5.2 | 1.3 |
No degree | 210 | 5.0 | 1.5 |
Records of 40 used passenger cars and 40 used pickup trucks (none used commercially) were randomly selected to investigate whether there was any difference in the mean time in years that they were kept by the original owner before being sold. For cars the mean was 5.3 years with standard deviation 2.2 years. For pickup trucks the mean was 7.1 years with standard deviation 3.0 years.
In previous years the average number of patients per hour at a hospital emergency room on weekends exceeded the average on weekdays by 6.3 visits per hour. A hospital administrator believes that the current weekend mean exceeds the weekday mean by fewer than 6.3 hours.
Construct the 99% confidence interval for the difference in the population means based on the following data, derived from a study in which 30 weekend and 30 weekday one-hour periods were randomly selected and the number of new patients in each recorded.
n | $\stackrel{-}{x}$ | s | |
---|---|---|---|
Weekends | 30 | 13.8 | 3.1 |
Weekdays | 30 | 8.6 | 2.7 |
A sociologist surveys 50 randomly selected citizens in each of two countries to compare the mean number of hours of volunteer work done by adults in each. Among the 50 inhabitants of Lilliput, the mean hours of volunteer work per year was 52, with standard deviation 11.8. Among the 50 inhabitants of Blefuscu, the mean number of hours of volunteer work per year was 37, with standard deviation 7.2.
A university administrator asserted that upperclassmen spend more time studying than underclassmen.
Test this claim against the default that the average number of hours of study per week by the two groups is the same, using the following information based on random samples from each group of students. Test at the 1% level of significance.
n | $\stackrel{-}{x}$ | s | |
---|---|---|---|
Upperclassmen | 35 | 15.6 | 2.9 |
Underclassmen | 35 | 12.3 | 4.1 |
An kinesiologist claims that the resting heart rate of men aged 18 to 25 who exercise regularly is more than five beats per minute less than that of men who do not exercise regularly. Men in each category were selected at random and their resting heart rates were measured, with the results shown.
n | $\stackrel{-}{x}$ | s | |
---|---|---|---|
Regular exercise | 40 | 63 | 1.0 |
No regular exercise | 30 | 71 | 1.2 |
Children in two elementary school classrooms were given two versions of the same test, but with the order of questions arranged from easier to more difficult in Version A and in reverse order in Version B. Randomly selected students from each class were given Version A and the rest Version B. The results are shown in the table.
n | $\stackrel{-}{x}$ | s | |
---|---|---|---|
Version A | 31 | 83 | 4.6 |
Version B | 32 | 78 | 4.3 |
The Municipal Transit Authority wants to know if, on weekdays, more passengers ride the northbound blue line train towards the city center that departs at 8:15 a.m. or the one that departs at 8:30 a.m. The following sample statistics are assembled by the Transit Authority.
n | $\stackrel{-}{x}$ | s | |
---|---|---|---|
8:15 a.m. train | 30 | 323 | 41 |
8:30 a.m. train | 45 | 356 | 45 |
In comparing the academic performance of college students who are affiliated with fraternities and those male students who are unaffiliated, a random sample of students was drawn from each of the two populations on a university campus. Summary statistics on the student GPAs are given below.
n | $\stackrel{-}{x}$ | s | |
---|---|---|---|
Fraternity | 645 | 2.90 | 0.47 |
Unaffiliated | 450 | 2.88 | 0.42 |
Test, at the 5% level of significance, whether the data provide sufficient evidence to conclude that there is a difference in average GPA between the population of fraternity students and the population of unaffiliated male students on this university campus.
In comparing the academic performance of college students who are affiliated with sororities and those female students who are unaffiliated, a random sample of students was drawn from each of the two populations on a university campus. Summary statistics on the student GPAs are given below.
n | $\stackrel{-}{x}$ | s | |
---|---|---|---|
Sorority | 330 | 3.18 | 0.37 |
Unaffiliated | 550 | 3.12 | 0.41 |
Test, at the 5% level of significance, whether the data provide sufficient evidence to conclude that there is a difference in average GPA between the population of sorority students and the population of unaffiliated female students on this university campus.
The owner of a professional football team believes that the league has become more offense oriented since five years ago. To check his belief, 32 randomly selected games from one year’s schedule were compared to 32 randomly selected games from the schedule five years later. Since more offense produces more points per game, the owner analyzed the following information on points per game (ppg).
n | $\stackrel{-}{x}$ | s | |
---|---|---|---|
ppg previously | 32 | 20.62 | 4.17 |
ppg recently | 32 | 22.05 | 4.01 |
Test, at the 10% level of significance, whether the data on points per game provide sufficient evidence to conclude that the game has become more offense oriented.
The owner of a professional football team believes that the league has become more offense oriented since five years ago. To check his belief, 32 randomly selected games from one year’s schedule were compared to 32 randomly selected games from the schedule five years later. Since more offense produces more offensive yards per game, the owner analyzed the following information on offensive yards per game (oypg).
n | $\stackrel{-}{x}$ | s | |
---|---|---|---|
oypg previously | 32 | 316 | 40 |
oypg recently | 32 | 336 | 35 |
Test, at the 10% level of significance, whether the data on offensive yards per game provide sufficient evidence to conclude that the game has become more offense oriented.
Large Data Sets 1A and 1B list the SAT scores for 1,000 randomly selected students. Denote the population of all male students as Population 1 and the population of all female students as Population 2.
http://www.gone.2012books.lardbucket.org/sites/all/files/data1A.xls
http://www.gone.2012books.lardbucket.org/sites/all/files/data1B.xls
Large Data Sets 1A and 1B list the GPAs for 1,000 randomly selected students. Denote the population of all male students as Population 1 and the population of all female students as Population 2.
http://www.gone.2012books.lardbucket.org/sites/all/files/data1A.xls
http://www.gone.2012books.lardbucket.org/sites/all/files/data1B.xls
Large Data Sets 7A and 7B list the survival times for 65 male and 75 female laboratory mice with thymic leukemia. Denote the population of all such male mice as Population 1 and the population of all such female mice as Population 2.
http://www.gone.2012books.lardbucket.org/sites/all/files/data7A.xls
http://www.gone.2012books.lardbucket.org/sites/all/files/data7B.xls
Z = 0.738, $\pm {z}_{0.025}=\pm 1.960$, do not reject H_{0} (no difference)
$Z=\text{\u2212}1.398$, $\text{\u2212}{z}_{0.10}=\text{\u2212}1.282$, reject H_{0} (more offense oriented)