Solutions

two proportions

matched or paired samples

single mean

independent group means, population standard deviations and/or variances unknown

two proportions

independent group means, population standard deviations and/or variances unknown

independent group means, population standard deviations and/or variances unknown

two proportions

The random variable is the difference between the mean amounts of sugar in the two soft drinks.

means

two-tailed

the difference between the mean life spans of whites and nonwhites

This is a comparison of two population means with unknown population standard deviations.

Check student’s solution.

1. Reject the null hypothesis.
2. p-value < 0.05
3. 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.

the difference in mean speeds of the fastball pitches of the two pitchers

–2.46

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.

Subscripts: 1 = Food, 2 = No Food
H₀: μ₁ ≤ μ₂
H_a: μ₁ > μ₂

Subscripts: 1 = Gamma, 2 = Zeta
H₀: μ₁ = μ₂
H_a: μ₁ ≠ μ₂

0.0062

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 OS₁ and OS₂.

0.1018

proportions

right-tailed

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.

Check student’s solution.

1. Reject the null hypothesis.
2. p-value < alpha
3. 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.

the mean difference of the system failures

0.0067

With a p-value 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.

0.0021

H₀: μ_d ≥ 0

H_a: μ_d < 0

0.0699

We decline to reject the null hypothesis. There is not sufficient evidence to support that the medication is effective.

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

1. H₀: μ₁ ≥ μ₂
2. H_a: μ₁ < μ₂
3. ${\overline{X}}_1–{\overline{X}}_2$ 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 percent significance level, there is sufficient evidence to conclude that the mean enrollment at four-year colleges is higher than at two-year colleges.

Subscripts: 1: mechanical engineering, 2: electrical engineering

1. H₀: µ₁ ≥ µ₂
2. H_a: µ₁ < µ₂
3. ${\overline{X}}_1-{\overline{X}}_2$ 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 percent significance level, there is insufficient evidence to conclude that the mean entry-level salaries of mechanical engineers is lower than that of electrical engineers.

1. H₀: µ₁ = µ₂
2. H_a: µ₁ ≠ µ₂
3. ${\overline{X}}_1-{\overline{X}}_2$ is the difference between the mean times for completing a lap in races and in practices.
4. t_20.32
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 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.

1. H₀: µ₁ = µ₂
2. H_a: µ₁ ≠ µ₂
3. is the difference between the mean times for completing a lap in races and in practices.
4. t_40.94
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 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.

c

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

• µ₁ = the mean price of a sociology text on the selected site
• µ₂ = 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: $H_0\text{:}{\mu }_1-{\mu }_2=0$, $H_a\text{:}{\mu }_1-{\mu }_2<0$, which can be expressed as $H_0\text{:}{\mu }_1={\mu }_2$, $H_a\text{:}{\mu }_1<{\mu }_2$.

Distribution for the test: Use t_df , because each sample has more than 30 observations, $df=n_1+n_2–2=33+33–2=64$.

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: .

Using a calculator with t_c = –2.55 and df = 64, the left-tailed p-value: 0.0066. Decision: Reject H₀. 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: H₀: μ₁ = μ₂, H_a: μ₁ < μ₂
The mean age of entering the industry in Canada is lower than the mean age in the United States.

Figure 10.20

Graph: left-tailed

p-value : 0.0151

Decision: Do not reject H₀.

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.

d

Subscripts: 1 = boys, 2 = girls

1. H₀: µ₁ ≤ µ₂
2. H_a: µ₁ > µ₂
3. The random variable is the difference in the mean auto insurance costs for boys and girls.
4. normal
5. test statistic: z = 2.50
6. p value: 0.0062
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 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 = non-hybrid sedans, 2 = hybrid sedans

1. H₀: µ₁ ≥ µ₂
2. H_a: µ₁ < µ₂
3. The random variable is the difference in the mean miles per gallon of nonhybrid sedans and hybrid sedans.
4. normal
5. test statistic: 6.36
6. p-value: 0
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 percent significance level, there is sufficient evidence to conclude that the mean miles per gallon of non-hybrid sedans is less than that of hybrid sedans.

1. H₀: µ_d = 0
2. H_a: µ_d < 0
3. The random variable X_d is the average difference between husband’s and wife’s satisfaction level.
4. t₉
5. test statistic: t = –1.86
6. p value: 0.0479
7. Check student’s solution
8. 1. Alpha: 0.05
   2. Decision: Reject the null hypothesis, but run another test.
   3. Reason for Decision: p value < alpha
   4. 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

1. H₀: p₁ = p₂
2. H_a: p₁ ≠ p₂
3. The random variable is the difference between the proportions of Hispanic students at Cabrillo College and Lake Tahoe College.
4. normal for two proportions
5. test statistic: 4.29
6. p-value: 0.00002
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: There is sufficient evidence to conclude that the proportions of Hispanic students at Cabrillo College and Lake Tahoe College are different.

a

Test: two independent sample proportions.

Random variable: p′₁ - p′₂

Distribution:
H₀: p₁ = p₂
H_a: p₁ ≠ p₂

The proportion of e-reader users is different for the 16- to 29-year-old users from that of the 30 and older users.

Graph: two-tailed

Figure 10.21

p-value : 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 e-reader users 16 to 29 years old is different from the proportion of e-reader 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, $df=n_1+n_2-2=33+33-2=64$.

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: $t_c=\frac{({\overline{X}}_1-{\overline{X}}_2)-0}{\sqrt{{\displaystyle \frac{{s_1}^2}{n_2}}+{\displaystyle \frac{{s_2}^2}{n_2}}}}=\frac{(74.64-111.56)-0}{\sqrt{{\displaystyle \frac{49.{36}^2}{33}}+{\displaystyle \frac{66.{90}^2}{33}}}}=-2.55$.

Using a calculator with $t_c=-2.55$ and $df=64$, the left-tailed $p$-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

1. H₀: p₁ ≤ p₂
2. H_a: p₁ > p₂
3. ${P^{\prime }}_1-{P^{\prime }}_2$ is the difference between the proportions of men and women who enjoy shopping for electronic equipment.
4. normal for two proportions
5. test statistic: 0.22
6. p-value: 0.4133
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 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.

1. H₀: p₁ = p₂
2. H_a: p₁ ≠ p₂
3. ${P^{\prime }}_1-{P^{\prime }}_2$ is the difference between the proportions of men and women that have at least one pierced ear.
4. normal for two proportions
5. test statistic: –4.82
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 percent significance level, there is sufficient evidence to conclude that the proportions of males and females with at least one pierced ear is different.

1. H₀: µ_d = 0
2. H_a: µ_d > 0
3. The random variable X_d is the mean difference in work times on days when eating breakfast and on days when not eating breakfast.
4. t₉
5. test statistic: 4.8963
6. p-value: 0.0004
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 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.

p-value = 0.1494

At the 5 percent significance level, there is insufficient evidence to conclude that the medication lowered cholesterol levels after 12 weeks.

b

c

Test: two matched pairs or paired samples (t-test)

Random variable: ${\overline{X}}_d$

Distribution: t₁₂

H₀: 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: right-tailed

p-value: 0.0004

Figure 10.22

Decision: Reject H₀.

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 (t-test)

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: H₀: 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: left-tailed.

Figure 10.23

p-value: 0.1207

Decision: Do not reject H₀.

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.

e

d

f

e

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.

a