Matched or Paired Samples (Optional)

A study was conducted to investigate the effectiveness of pain-reducing medication. Results for randomly selected subjects are shown in table-2345. A lower score indicates less pain. The before value is matched to an after value, and the differences are calculated. The differences have a normal distribution. Are the sensory measurements, on average, lower after the medication? Test at a 5 percent significance level.

Corresponding before and after values form matched pairs. (Calculate after – before.)

The data for the test are the differences: {0.2, –4.1, –1.6, –1.8, –3.2, –2, –2.9, –9.6}

The sample mean and sample standard deviation of the differences are: $d_{}=\mathrm{–3.13}$ and $s_d=2.91$
Verify these values.

Let ${\mu }_d$ be the population mean for the differences. We use the subscript $d$ to denote differences.

Random variable: ${\overline{X}}_d$ = the mean difference of the sensory measurements.

H₀: μ_d ≥ 0

The null hypothesis is zero or positive, meaning that there is the same or more pain felt after taking the medication. That means the subject shows no improvement. μ_d is the population mean of the differences.

H_a: μ_d < 0

The alternative hypothesis is negative, meaning there is less pain felt after taking the medication. That means the subject shows improvement. The score should be lower after taking the medication, so the difference ought to be negative to indicate improvement.

Distribution for the test: The distribution is a Student’s t with df = n – 1 = 8 – 1 = 7. Use t₇. Note —that the test is for a single population mean.

Calculate the p-value using the Student’s-t distribution: p-value = 0.0095

Graph:

View original figure (external; image not reproduced in this text edition)

Original figure description: Normal distribution curve of the average difference of sensory measurements with values of -3.13 and 0. A vertical upward line extends from -3.13 to the curve, and the p-value is indicated in the area to the left of this value.

${\overline{X}}_d$ is the random variable for the differences.

The sample mean and sample standard deviation of the differences are as follows:

${\overline{x}}_d$ = –3.13

${\overline{s}}_d$ = 2.91

Compare α and the p-value: α = 0.05 and p-value = 0.0095. α > p-value.

Make a decision: Since α > p-value, reject H₀. This means that μ_d < 0 and there is improvement.

Conclusion: At a 5 percent level of significance, from the sample data, there is sufficient evidence to conclude that the sensory measurements, on average, are lower after taking the medication. The medication appears to be effective in reducing pain.

What is the random variable?

the mean difference of the system failures

State the null and alternative hypotheses.

What is the p-value?

0.0067

Draw the graph of the p-value.

What conclusion can you draw about the software patch?

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.

State the null and alternative hypotheses.

What is the p-value?

0.0021

What is the sample mean difference?

Draw the graph of the p-value.

View original figure (external; image not reproduced in this text edition)

Original figure description: This is a normal distribution curve with mean equal to zero. The values 0 and 1.67 are labeled on the horiztonal axis. A vertical line extends from 1.67 to the curve. The region under the curve to the right of the line is shaded to represent p-value = 0.0021.

What conclusion can you draw about the juggling class?

State the null and alternative hypotheses.

H₀: μ_d ≥ 0

H_a: μ_d < 0

What is the test statistic?

What is the p-value?

0.0699

What is the sample mean difference?

What is the conclusion?

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

Ten individuals went on a low-fat diet for 12 weeks to lower their cholesterol. The data are recorded in fs-idp16051952. Do you think that their cholesterol levels were significantly lowered?

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.

The appropriate hypotheses are

1. H₀: p_t < p_ut and H_a: p_t ≥ p_ut
2. H₀: p_t ≤ p_ut and H_a: p_t > p_ut
3. H₀: p_t = p_ut and H_a: p_t ≠ p_ut
4. H₀: p_t = p_ut and H_a: p_t < p_ut

If the p-value is 0.0062, what is the conclusion? Use α = 0.05.

1. The method has no effect.
2. There is sufficient evidence to conclude that the method reduces the proportion of patients who develop the disease after four years.
3. There is sufficient evidence to conclude that the method increases the proportion of patients who develop the disease after four years.
4. There is insufficient evidence to conclude that the method reduces the proportion of patients who develop the disease after four years.

b

The distribution for the test is

1. t₅
2. t₆
3. N(−10.2, 8.4)
4. N(−10.2, $\frac{8.4}{\sqrt{6}}$)

If α = 0.05, the p-value and the conclusion are

1. 0.0014; There is sufficient evidence to conclude that the blood pressure decreased after the training.
2. 0.0014; There is sufficient evidence to conclude that the blood pressure increased after the training.
3. 0.0155; There is sufficient evidence to conclude that the blood pressure decreased after the training.
4. 0.0155; There is sufficient evidence to conclude that the blood pressure increased after the training.

c

A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective. She takes four new students. She records their 18-hole scores before learning the technique and then after having taken her class. She conducts a hypothesis test. The data are as follows.

The correct decision is

1. reject H₀.
2. do not reject H₀.

A local research group is studying a chronic disease. They believe the number of cases of the disease is higher in 2013 than in 2012 in the southern United States. The group compared the estimates of new cases by southern state in 2012 and 2013. The results are in fs-idp20856160.

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

View original figure (external; image not reproduced in this text edition)

Original figure description: This is a normal distribution curve with mean equal to zero. A vertical line near the tail of the curve to the right of zero extends from the axis to the curve. The region under the curve to the right of the line is shaded representing p-value = 0.0004.

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.

A traveler wanted to know if the prices of hotels are different in the 10 cities that he visits the most often. The list of the cities with the corresponding prices for his two favorite hotel chains is in fs-idp126693504. Test at the 1 percent level of significance.

A politician asked his staff to determine whether the underemployment rate in the Northeast decreased from 2011 to 2012. The results are in fs-idp10131504.

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.

View original figure (external; image not reproduced in this text edition)

Original figure description: This is a normal distribution curve with mean equal to zero. A vertical line near the tail of the curve to the right of zero extends from the axis to the curve. The region under the curve to the right of the line is shaded representing p-value = 0.1207.

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.

A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. The population standard deviations are two pounds and three pounds, respectively. Of interest is whether the liquid diet yields a higher mean weight loss than the powder diet.

A new chocolate bar is taste-tested on consumers. Of interest is whether the proportion of children who like the new chocolate bar is greater than the proportion of adults who like it.

e

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 9 males and 16 females.

A football league reported that the mean number of touchdowns per game was five. A study is done to determine if the mean number of touchdowns has decreased.

d

A study is done to determine if students in the California state university system take longer to graduate than students enrolled in private universities. One hundred students from both the California state university system and private universities are surveyed. From years of research, it is known that the population standard deviations are 1.5811 years and 1 year, respectively.

According to a doctor’s magazine, 75 percent of senior citizens think that yearly checkups are very important. A study is done to verify this.

f

According to a recent study, U.S. companies have a mean maternity leave of six weeks.

A recent survey showed an increase in use of prescription medication among local senior citizens as compared to the national percent. Suppose that a survey of 100 local senior citizens and 100 national senior citizens is conducted to see if the proportion of prescription medication use is higher locally than nationally.

e

A new SAT study course is tested on 12 individuals. Pre-course and post-course scores are recorded. Of interest is the mean increase in SAT scores. The following data are collected:

According to a statistics college professor, 68 percent of his students pass the final exam. A graduate researcher designs a study to determine if this claim is true.

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.

Lesley E. Tan investigated the relationship between left-handedness versus right-handedness and motor competence in preschool children. Random samples of 41 left-handed preschool children and 41 right-handed preschool children were given several tests of motor skills to determine if there is evidence of a difference between the children based on this experiment. The experiment produced the means and standard deviations shown in fs-idp128816944. Determine the appropriate test and best distribution to use for that test.

1. Two independent means, normal distribution
2. Two independent means, Student’s t-distribution
3. Matched or paired samples, Student’s t-distribution
4. Two population proportions, normal distribution

A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective. She takes four new students. She records their 18-hole scores before learning the technique and after having taken her class. She conducts a hypothesis test. The data are shown in fs-idp136885584.

This is

1. a test of two independent means.
2. a test of two proportions.
3. a test of a single mean.
4. a test of a single proportion.

a