 Reject the null hypothesis.
 pvalue < 0.05
 There is not enough evidence at the 5 percent level of significance to support the claim that life expectancy in the 1900s is different between whites and nonwhites.
At the 1 percent significance level, we can reject the null hypothesis. There is sufficient data to conclude that the mean speed of Rodriguez’s fastball is faster than Wesley’s.
There is sufficient evidence to reject the null hypothesis. The data support that the melting point for Alloy Zeta is different from the melting point of Alloy Gamma.
P′_{OS1} – P′_{OS2} = difference in the proportions of phones that had system failures within the first eight hours of operation with OS_{1} and OS_{2}.
The random variable is the difference in proportions (percents) of the populations that are of two or more races in Nevada and North Dakota.
Our sample sizes are much greater than five each, so we use the normal for two proportions distribution for this hypothesis test.
 Reject the null hypothesis.
 pvalue < alpha
 At the 5 percent significance level, there is sufficient evidence to conclude that the proportion (percent) of the population that is of two or more races in Nevada is statistically higher than that in North Dakota.
With a pvalue 0.0067, we can reject the null hypothesis. There is enough evidence to support that the software patch is effective in reducing the number of system failures.
We decline to reject the null hypothesis. There is not sufficient evidence to support that the medication is effective.
Subscripts: 1: twoyear colleges, 2: fouryear colleges
 H_{0}: μ_{1} ≥ μ_{2}
 H_{a}: μ_{1} < μ_{2}
 ${\overline{X}}_{1}\u2013{\overline{X}}_{2}$ is the difference between the mean enrollments of the twoyear colleges and the fouryear colleges.
 Student’s t
 test statistic: 0.2480
 pvalue: 0.4019
 Check student’s solution.

 Alpha: 0.05
 Decision: Do not reject.
 Reason for Decision: pvalue > alpha
 Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the mean enrollment at fouryear colleges is higher than at twoyear colleges.
Subscripts: 1: mechanical engineering, 2: electrical engineering
 H_{0}: µ_{1} ≥ µ_{2}
 H_{a}: µ_{1} < µ_{2}
 ${\overline{X}}_{1}{\overline{X}}_{2}$ is the difference between the mean entrylevel salaries of mechanical engineers and electrical engineers.
 t_{108}
 test statistic: t = –0.82
 pvalue: 0.2061
 Check student’s solution.

 Alpha: 0.05
 Decision: Do not reject the null hypothesis.
 Reason for Decision: pvalue > alpha
 Conclusion: At the 5 percent significance level, there is insufficient evidence to conclude that the mean entrylevel salaries of mechanical engineers is lower than that of electrical engineers.
 H_{0}: µ_{1} = µ_{2}
 H_{a}: µ_{1} ≠ µ_{2}
 ${\overline{X}}_{1}{\overline{X}}_{2}$ is the difference between the mean times for completing a lap in races and in practices.
 t_{20.32}
 test statistic: –4.70
 pvalue: 0.0001
 Check student’s solution.

 Alpha: 0.05
 Decision: Reject the null hypothesis.
 Reason for Decision: pvalue < alpha
 Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the mean time for completing a lap in races is different from that in practices.
 H_{0}: µ_{1} = µ_{2}
 H_{a}: µ_{1} ≠ µ_{2}
 is the difference between the mean times for completing a lap in races and in practices.
 t_{40.94}
 test statistic: –5.08
 pvalue: zero
 Check student’s solution.

 Alpha: 0.05
 Decision: Reject the null hypothesis.
 Reason for Decision: pvalue < alpha
 Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the mean time for completing a lap in races is different from that in practices.
Test: two independent sample means, population standard deviations unknown.
 µ_{1} = the mean price of a sociology text on the selected site
 µ_{2} = the mean price of a math/science text on the selected site
Random variable: ${\overline{X}}_{1}{\overline{X}}_{2}$ = the difference in the sample mean textbook price between sociology texts and math/science texts.
Hypotheses: ${H}_{0}\text{:}\phantom{\rule{0.3em}{0ex}}{\mu}_{1}{\mu}_{2}=0$, $\phantom{\rule{0.6em}{0ex}}{H}_{a}\text{:}\phantom{\rule{0.3em}{0ex}}{\mu}_{1}{\mu}_{2}<0$, which can be expressed as ${H}_{0}\text{:}\phantom{\rule{0.3em}{0ex}}{\mu}_{1}={\mu}_{2}$, ${H}_{a}\text{:}\phantom{\rule{0.3em}{0ex}}{\mu}_{1}<{\mu}_{2}$.
Distribution for the test: Use t_{df }, because each sample has more than 30 observations, $df=\phantom{\rule{0.3em}{0ex}}{n}_{1}+{n}_{2}\u20132=33+33\u20132=64$.
Estimate the critical value on the ttable using the nearest available degrees of freedom, 60. The critical value, 2.660, is found in the .0005 column.
Calculate the test statistic: .
Using a calculator with t_{c} = –2.55 and df = 64, the lefttailed pvalue: 0.0066. Decision: Reject H_{0}. Conclusion: At the 1% level of significance, from the sample data, there is sufficient evidence to conclude that the mean price of sociology textbooks is less than the mean price of textbooks for math/science.
Distribution: H_{0}: μ_{1} = μ_{2}, H_{a}: μ_{1} < μ_{2}
The mean age of entering the industry in Canada is lower than the mean age in the United States.
Graph: lefttailed
pvalue : 0.0151
Decision: Do not reject H_{0}.
Conclusion: At the 1 percent level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of entering the industry in Canada is lower than the mean age in the United States.
Subscripts: 1 = boys, 2 = girls
 H_{0}: µ_{1} ≤ µ_{2}
 H_{a}: µ_{1} > µ_{2}
 The random variable is the difference in the mean auto insurance costs for boys and girls.
 normal
 test statistic: z = 2.50
 p value: 0.0062
 Check student’s solution.

 Alpha: 0.05
 Decision: Reject the null hypothesis.
 Reason for Decision: p value < alpha
 Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the mean cost of auto insurance for teenage boys is greater than that for girls.
Subscripts: 1 = nonhybrid sedans, 2 = hybrid sedans
 H_{0}: µ_{1} ≥ µ_{2}
 H_{a}: µ_{1} < µ_{2}
 The random variable is the difference in the mean miles per gallon of nonhybrid sedans and hybrid sedans.
 normal
 test statistic: 6.36
 pvalue: 0
 Check student’s solution.

 Alpha: 0.05
 Decision: Reject the null hypothesis.
 Reason for decision: p value < alpha
 Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the mean miles per gallon of nonhybrid sedans is less than that of hybrid sedans.
 H_{0}: µ_{d} = 0
 H_{a}: µ_{d} < 0
 The random variable X_{d} is the average difference between husband’s and wife’s satisfaction level.
 t_{9}
 test statistic: t = –1.86
 p value: 0.0479
 Check student’s solution
 Alpha: 0.05
 Decision: Reject the null hypothesis, but run another test.
 Reason for Decision: p value < alpha
 Conclusion: This is a weak test because alpha and the p value are close. However, there is insufficient evidence to conclude that the mean difference is negative.
Subscripts: 1 = Cabrillo College, 2 = Lake Tahoe College
 H_{0}: p_{1} = p_{2}
 H_{a}: p_{1} ≠ p_{2}
 The random variable is the difference between the proportions of Hispanic students at Cabrillo College and Lake Tahoe College.
 normal for two proportions
 test statistic: 4.29
 pvalue: 0.00002
 Check student’s solution.

 Alpha: 0.05
 Decision: Reject the null hypothesis.
 Reason for decision: pvalue < alpha
 Conclusion: There is sufficient evidence to conclude that the proportions of Hispanic students at Cabrillo College and Lake Tahoe College are different.
Test: two independent sample proportions.
Random variable: p′_{1}  p′_{2}
Distribution:
H_{0}: p_{1} = p_{2}
H_{a}: p_{1} ≠ p_{2}
The proportion of ereader users is different for the 16 to 29yearold users from that of the 30 and older users.
Graph: twotailed
pvalue : 0.0033
Decision: Reject the null hypothesis.
Conclusion: At the 5 percent level of significance, from the sample data, there is sufficient evidence to conclude that the proportion of ereader users 16 to 29 years old is different from the proportion of ereader users 30 and older.
Test: two independent sample means, population standard deviations unknown.
${\mu}_{1}$ = the the mean price of a sociology text on the selected site.
${\mu}_{2}$ = the mean price of a math/science text on the selected site.
Random variable: $\overline{{X}_{1}}\overline{{X}_{1}}$ = the difference in the sample mean textbook price between sociology texts and math/science texts.
Hypotheses: ${H}_{0}:{\mu}_{1}{\mu}_{2}=0,{H}_{a}:{\mu}_{1}{\mu}_{2}{\mu}_{2}$ which can be expressed as $\mathrm{H}0\mathrm{s}:\mathrm{\mu}1\mathrm{\mu}2,\mathrm{Ha}\mathrm{\mu}1\mathrm{\mu}2$.
Distribution for the test: Use ${t}_{df}$; because each sample has more than 30 observations, $df={n}_{1}+{n}_{2}2=33+332=64$.
Estimate the critical value on the $t$table using the nearest available degrees of freedom, 60. The critical value, 2.660, is found in the .0005 column.
Calculate the test statistic: ${t}_{c}=\frac{({\overline{X}}_{1}{\overline{X}}_{2})0}{\sqrt{{\displaystyle \frac{{{s}_{1}}^{2}}{{n}_{2}}}+{\displaystyle \frac{{{s}_{2}}^{2}}{{n}_{2}}}}}=\frac{(74.64111.56)0}{\sqrt{{\displaystyle \frac{49.{36}^{2}}{33}}+{\displaystyle \frac{66.{90}^{2}}{33}}}}=2.55$.
Using a calculator with ${t}_{c}=2.55$ and $df=64$, the lefttailed $p$value: Decision: Reject ${H}_{0}$. Conclusion: At the 1% level of significance, from the sample data, there is sufficient evidence to conclude that the mean price of sociology textbooks is less than the mean price of textbooks for math/science.
Subscripts: 1: men; 2: women
 H_{0}: p_{1} ≤ p_{2}
 H_{a}: p_{1} > p_{2}
 ${{P}^{\prime}}_{1}{{P}^{\prime}}_{2}$ is the difference between the proportions of men and women who enjoy shopping for electronic equipment.
 normal for two proportions
 test statistic: 0.22
 pvalue: 0.4133
 Check student’s solution.

 Alpha: 0.05
 Decision: Do not reject the null hypothesis.
 Reason for Decision: pvalue > alpha
 Conclusion: At the 5 percent significance level, there is insufficient evidence to conclude that the proportion of men who enjoy shopping for electronic equipment is more than the proportion of women.
 H_{0}: p_{1} = p_{2}
 H_{a}: p_{1} ≠ p_{2}
 ${{P}^{\prime}}_{1}{{P}^{\prime}}_{2}$ is the difference between the proportions of men and women that have at least one pierced ear.
 normal for two proportions
 test statistic: –4.82
 pvalue: zero
 Check student’s solution.

 Alpha: 0.05
 Decision: Reject the null hypothesis.
 Reason for Decision: pvalue < alpha
 Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the proportions of males and females with at least one pierced ear is different.
 H_{0}: µ_{d} = 0
 H_{a}: µ_{d} > 0
 The random variable X_{d} is the mean difference in work times on days when eating breakfast and on days when not eating breakfast.
 t_{9}
 test statistic: 4.8963
 pvalue: 0.0004
 Check student’s solution.

 Alpha: 0.05
 Decision: Reject the null hypothesis.
 Reason for Decision: pvalue < alpha
 Conclusion: At the 5 percent level of significance, there is sufficient evidence to conclude that the mean difference in work times on days when eating breakfast and on days when not eating breakfast has increased.
pvalue = 0.1494
At the 5 percent significance level, there is insufficient evidence to conclude that the medication lowered cholesterol levels after 12 weeks.
Test: two matched pairs or paired samples (ttest)
Random variable: ${\overline{X}}_{d}$
Distribution: t_{12}
H_{0}: μ_{d} = 0 H_{a}: μ_{d} > 0
The mean of the differences of new female breast cancer cases in the south between 2013 and 2012 is greater than zero. The estimate for new female breast cancer cases in the south is higher in 2013 than in 2012.
Graph: righttailed
pvalue: 0.0004
Decision: Reject H_{0}.
Conclusion: At the 5 percent level of significance, from the sample data, there is sufficient evidence to conclude that there was a higher estimate of new female breast cancer cases in 2013 than in 2012.
Test: matched or paired samples (ttest)
Difference data: {–0.9, –3.7, –3.2, –0.5, 0.6, –1.9, –0.5, 0.2, 0.6, 0.4, 1.7, –2.4, 1.8}
Random Variable: ${\overline{X}}_{d}$
Distribution: H_{0}: μ_{d} = 0 H_{a}: μ_{d} < 0
The mean of the differences of the rate of underemployment in the northeastern states between 2012 and 2011 is less than zero. The underemployment rate went down from 2011 to 2012.
Graph: lefttailed.
pvalue: 0.1207
Decision: Do not reject H_{0}.
Conclusion: At the 5 percent level of significance, from the sample data, there is not sufficient evidence to conclude that there was a decrease in the underemployment rates of the northeastern states from 2011 to 2012.
f
The graduate researcher will be comparing a sample proportion to a population proportion or claim. Thus, the study includes the hypothesis test of a single proportion. A two proportion hypothesis test compares two sample proportions.