DPH 623 Final

Statistics

A body of techniques and procedures dealing with the collection, organization, interpretation, analysis and presentation of information that can be stated numerically.

Parameter

A numerical value computed from a population is called a ________

Skewed to the left

A skewed distribution with a tail that stretches left, toward the smaller values.

Body mass index (BMI) is equal to "weight in kilograms" divided by "height in meters squared." A study by the National Center for Health Statistics suggested that women between the ages of 20 and 29 in the United States have a mean BMI of 26.8 with a standard deviation of 7.4. Let us assume that these BMIs are normally distributed. a) A BMI of 30 or greater is classified as being overweight. What proportion of women in this age range are overweight according to this definition? b) A BMI less than 18.5 is considered to be underweight. What proportion of women are underweight? Show your work.

A) z=30-26.8/7.4 = 0.4324. Find 0.43 in the Z table, which equals 0.6664. 1-0.664 = 0.3336. B) z=18.5-26.8/7.4 = -1.12. Find -1.12 in the Z table, which equals 0.1314

skewed right distribution

the tail to the right of the peak is longer than the tail to the left of the peak

Bar charts are best used to display the information on data collected from non-measurable qualitative variables.

true

Box plots are the best way to graphically represent the data obtained from research studies that focus on measuring the outcomes for variables considered to exist on a quantitative continuum.

true

Suppose that scores on the biological sciences section of the Medical College Admissions Test (MCAT) are Normally distributed with a mean of 9.2 and standard deviation of 2.2. Successful applicants to become medical students had a mean score of 10.8 on this portion of the test. What percentage of applicants had a score of 10.8 or greater? Show your work.

z= 10.8-9.2 / 2.2 = 07273 0.73 in the Z table = 0.7673 1-0.7673 = 0.2327 23.3 % of applicants had a score of 10.8 or greater.

What is the mean of the Standard Normal distribution? [ ]

zero

The data shown below are the times (in weeks) required to complete 14 consulting projects. 1 2 6 12 13 13 13 15 16 17 18 19 20 26 The number of outliers in a modified boxplot of this data set is ______

2

One student in this class has the following scores for the first four quizzes: 7, 8, 10, 3 The standard deviation for this sample of four quiz scores is ____

2.94

A statistical program output is shown below for a sample of n = 142 hourly employees on the variable X = job performance rating on a scale of 0.0 to 10.0. Use this output to answer the questions below the output. Descriptive Statistics: Rating Variable N Mean StDev Minimum Q1 Median Q3 Maximum Rating 142 7.91 1.56 3.40 7.10 8.30 9.00 10.00 The percentage of performance ratings that are greater than 9.0 is ____

25%

One student in this class has the following scores for the first four quizzes: 7, 8, 10, 3 The mean for this sample of four quiz scores is ____

7.0

What is the z-value for the first quartile value?

-0.675

Which of the following hypothetical scenarios best describes where approximation by the binomial distribution is most acceptable?

-Rolling three doubles in a game of monopoly and going straight to jail -Correct! Flipping a coin six times, and getting 4 heads T-he probability of rolling two six-sided die, and getting a 7 -Finding the probability of selecting a red ball from a vase containing red, green, and yellow balls after selecting two yellow balls without replacement.

What is the probability (Area Under Curve) of the following: Pr(Z > 1.08)?

0.1401

What is the cumulative probability for z = - 1.07 in the table?

0.1423

What is the probability (Area Under Curve) of the following: Pr(- 2.13 ≤ Z ≤ 1.57)?

0.9257

Hyperlipidemia in children has been hypothesized to be related to high cholesterol in their parents. The following data were collected on parents and children. Child Both Parents Hyperlipidemic One Parent Hyperlipidemic Neither Parent Hyperlipidemic Not hyperlipidemic 13 34 83 Hyperlipidemic 45 42 6 What is the probability that one or both parents are hyperlipidemic? What is the probability that the child and both parents are hyperlipidemic? What is the probability that a child is hyperlipidemic if neither parent is hyperlipidemic? What is the probability that a child is hyperlipidemic if both parents are hyperlipidemic?

1) 42 + 45 = 87. 87/223 = 0.39 2) 45 /223 = 0.20 3) 6 / 223 = 0.027 4) 45/223 = 0.20

State the null and the alternative hypothesis for the following statement. Hint: use the statement H0: ........... Ha: ........... The mean number of years Americans work before retiring is 34. At most 60% of Americans vote in presidential elections. The mean starting salary for San Jose State University graduates is at least $100,000 per year. Twenty-nine percent of high school seniors get drunk each month. Fewer than 5% of adults ride the bus to work in Los Angeles. The mean number of cars a person owns in her lifetime is not more than ten. About half of Americans prefer to live away from cities, given the choice. Europeans have a mean paid vacation each year of six weeks. The chance of developing breast cancer is under 11% for women. Private universities' mean tuition cost is more than $20,000 per year.

1) : H0: The null hypothesis for the mean number of years Americans work before retiring is 34. The alternative hypothesis, Ha, for the mean number of years Americans work before retiring, is not 34. 2) The null hypothesis is the percentage of Americans who vote in presidential elections is 60%. The alternative hypothesis for the percentage of Americans who vote in presidential elections is greater than 60%. 3) The null hypothesis for the mean starting salary for San Jose State University graduates is equal to or less than $100,000 per year. The alternative hypothesis for the mean starting salary for San Jose State University graduates is greater than $100,000 per year. 4) The null hypothesis is the percentage of high school seniors who get drunk each month is 29%. The alternative hypothesis for the percentage of high school seniors who get drunk each month is not 29%. 5) The null hypothesis for the percentage of adults who ride the bus to work in Los Angeles is 5% or greater. The alternative hypothesis for the percentage of adults who ride the bus to work in Los Angeles is less than 5%. 6) The null hypothesis for the mean number of cars a person owns in her lifetime is ten or less. The alternative hypothesis for the mean number of cars a person owns in her lifetime is more than ten. 7) The null hypothesis states that about half of Americans prefer to live away from cities. The alternative hypothesis suggests that about half of Americans do not prefer to live away from cities. 8) The null hypothesis states that Europeans have a mean paid vacation each year of six weeks. The alternative hypothesis suggests that Europeans' mean paid vacation each year is not six weeks. 9) The null hypothesis states that the chance of developing breast cancer is equal to or under 11% for women. The alternative hypothesis suggests that the

If the null hypothesis is BLANK, we conclude that the alternative hypothesis is BLANK. If the null is BLANK, we conclude that the null may be BLANK.

1) rejected 2) true 3) not rejected 4) true

A statistical program output is shown below for a sample of n = 142 hourly employees on the variable X = job performance rating on a scale of 0.0 to 10.0. Use this output to answer the questions below the output. Descriptive Statistics: Rating Variable N Mean StDev Minimum Q1 Median Q3 Maximum Rating 142 7.91 1.56 3.40 7.10 8.30 9.00 10.00 The interquartile range for the performance ratings is ____

1.9

What is the median blood glucose level of the following data set collected from 8 individuals: 89, 95, 99, 102, 107, 108, 111, and 119?

104.5

The data shown below are the times (in weeks) required to complete 14 consulting projects. 1 2 6 12 13 13 13 15 16 17 18 19 20 26 The first quartile for this data set is ____

12

The data shown below are the times (in weeks) required to complete 14 consulting projects. 1 2 6 12 13 13 13 15 16 17 18 19 20 26 The median for this data set is ____ (A) 13 (B) 14 (C) 15 (D) 15.5

14

The data shown below are the times (in weeks) required to complete 14 consulting projects. 1 2 6 12 13 13 13 15 16 17 18 19 20 26 The value of Q3 for this data set is ____

18

Given the table below from a bipolar disorder study among young people, please find the probabilities by showing the steps. Family History of Mood Disorders Age =< 18 Age > 18 Total Negative 28 35 63 Bipolar disorder 19 38 57 Unipolar 41 44 85 Unipolar and bipolar 53 60 113 Total 141 177 318 What is the probability of a person who will be 18 years old or younger? What is the probability of a person who will be bipolar disorder? What is the probability of a person who will be 18 years old or younger and bipolar disorder? What is the probability of a person who will be 18 years old or younger or or bipolar disorder? What is the probability of a person who will not be bipolar disorder?

1: 0.44 141/318 = 0.44 2: 0.18 57/318 = 0.179 = 0.18 3: 0.06 19/318 = 0.059 = 0.06 4: 0.62 141 + 57 = 198. 198/318 = 0.62 5: 0.82 318-57 = 261. 261/318 = 0.82

A medical research team wished to evaluate a proposed screening test for Alzheimer's disease. The test was given to a random sample of 450 patients with Alzheimer's disease and an independent random sample of 500 patients without symptoms of the disease. The two samples were drawn from populations of subjects who were 65 years of age or older. The results are as follows: Alzheimer's Diagnosis Test Result Yes (D+) No (D-) Total Positive (T+) 436 5 441 Negative (T-) 14 495 509 Total 450 500 950 Given that 11.3 percent of the U.S. population aged 65 and over have Alzheimer's disease Solve the questions below showing your steps. estimate the sensitivity of the test. Estimate specificity of the test. compute the predictive value positive of the test. Compute the predictive value negative of the test.

1: 436/450 = 0.968 = 0.97 A is 436. C is 14. A + C = 450. 2: 495/500 = 0.99 D is 495. B is 5. D + B = 500. 3: 0.923. The sensitivity of the test is 0.968. 436/450. Positive test results without symptoms of Alzheimer's = 5. 5/500 = 0.1 Estimate of P(D) = .113. The percentage of the US population aged 65 and over who have Alzheimer's is 11.3 percent. Two independent samples were drawn from two different populations. It was not drawn from the relevant general population. (0.9689)(0.113)/(0.9689)(0.113) + (0.1)(1-0.113) = 0.923 (0.9689)(0.113) = 0.109 (0.9689)(0.113) + (0.01)(1-0.113) = 0.118 (0.109) / (0.118) = 0.9237 4: 0.996 The specificity of the test is 0.99. 450/500. 14/450 = 0.031 (0.99)(1-0.113) / (0.99)(1-0.113) + (0.031)(.113) = 0.996 (0.99)(1-0.113) = 0.87813 (0.99)(1-0.113) + (0.031)(0.113) = 0.881633 (0.87813) / (0.881633) = 0.996

The average number of errors on a page of a certain magazine is 0.2. What is the probability that the next page (or a randomly selected page) you read contains? (Hint: Poisson Distribution) 0 (zero) error? 2 or more errors? What is the average error per page? Find standard deviation of the number of errors

1: X=0, e=2.7183, mean/average = 0.2 e^-0.2 x 0.2^0 / 0! = 0.8187 2: X = 2 or more, e =2.7183, mean/average = 0.2 1 - P (X=0) and P (X=1) P X=0 = 0.8187 P X=1, e^-0.2 x 0.2^1 / 1! = 0.1637 0.8187 + 0.1637 = 0.9824 1-0.9824 = 0.0176 3: The average error per page is 0.2. 4: The square root of 0.2 = 0.4472. The standard deviation of the number of errors is 0.4472.

The number of viewers of a TV show per week has a mean of 29 million with a standard deviation of 5 million. Assume that, the number of viewers of that show follows a normal distribution. Hint: Use the z-table What is the probability that, next week's show will Have between 30 and 34 million viewers (hint: X1≤X≤X2) Have at least 23 million viewers? Exceed 40 million viewers?

1: Z1 = (30-29) / 5 = 0.2, The Z score = 0.5793. Z2 = (34-29) / 5 = 1 The Z score is 0.8413. 0.8413-0.5793 = 0.262 2: (23-29) / 5 = -1.2 = 0.11507. 1-0.11507 = 0.8849 3: (40-29) / 5 = 2.2 = 0.98610. 1-0.98610 = 0.0139.

Kyle's doctor told him that the z-score for his systolic blood pressure is 1.75. Which of the following is the best interpretation of this standardized score? The systolic blood pressure (given in millimeters) of males has an approximately normal distribution with mean μ=125 and standard deviation σ=14. If X= a systolic blood pressure score then X∼N(125,14). 1) Which answer is correct? Show your work A) Kyle's systolic blood pressure is 175. B) Kyle's systolic blood pressure is 1.75 times the average blood pressure of men his age. C) Kyle's systolic blood pressure is 1.75 above the average systolic blood pressure of men his age. D) Kyles's systolic blood pressure is 1.75 standard deviations above the average systolic blood pressure for men. 2) Calculate Kyle's blood pressure.

A: (175-125) / (14) = 3.571. The Z score is 1.75. FALSE B: A Z score does not measure "times the average." It is a measure of the standard deviations above or below a mean. FALSE C: A Z score is the number of standard deviations a value is above or below the mean. We are not adding the Z score (1.75) to the average. D: A Z score is the number of standard deviations a value is above or below the mean. Kyle's systolic blood pressure is 1.75 standard deviations above the average systolic blood pressure for men. TRUE. 2) mean = 125, z score = 1.75, standard deviation = 14 1.75 = (X - 125) / (14) 1.75 (14) = (X - 125) 24.5 = X - 125 24.5 + 125 = X 149.5 = X

The 2010 U.S. Census found the chance of a household being a certain size. The data is in the table below. Table: Household Size from U.S. Census of 2010 Size of household 1 2 3 4 5 6 7 or more Probability 26.7% 33.6% 15.8% 13.7% 6.3% 2.4% 1.5% State the random variable (hint: the x). Is it unusual for a household to have six people in the family? (hint: x≤6) If you did come upon many families that had six people in the family, what would you think? (explain) Is it unusual for a household to have four people in the family? (hint: x≤4, x≥4)

A: Size of household B: Yes, it is unusual for a household to have six people in the family. The probability is out of 100. Out of 100%, the probability for a household of 6 is only 2.4%. C: I would think that the trial gathered data from a sample that was not representative of the population. D: No, I do not think it is unusual for a household to have four people. 13.7% is a higher probability than households with 5, 6, and 7 or more. Households with three people have 15.8%.

Statistics can be categorized as either descriptive statistics or inferential statistics. Of the following research statements, which is the most far from others?

According to Census, we have around 70% Caucasian, 16% Hispanic, 12% African-American, and others in 2000

After analyzing data, Amy found that her data does not represent population since some characteristics are so different from those of previous researches. If a sample does not representing population, then the data has some ________.

Biases

Descriptive statistics are broken down into measures of blank and blank measures.

Central tendency and dispersion

What type of variable is X = body temperature?

Continuous

Amy should collect a sample data size of 50, but she does not have enough time. So, she collected those from her three different classes' students. What is this sample method?

Convenience Sampling

We want to test if college students take less than five years to graduate from college, on average. State the null and alternative hypotheses in terms of H0 and Ha

H0= μ ≥ 5. College students take greater than or equal to five years to graduate from college, on average. HA= μ < 5. College students take less than five years to graduate from college, on average.

Which one of the following is not part of the five-number summary?

Mean

What type of variable is X = student ID (example: 903650)?

Nominal

What type of variable is Y = subject identification number?

Nominal

When a new drug is created, the pharmaceutical company must subject it to testing before receiving the necessary permission from the Food and Drug Administration (FDA) to market the drug. Suppose the null hypothesis is "the drug is unsafe." What is Type II Error?

Not to conclude the drug is safe when, in fact, it is safe.

Of the following, which is NOT true about sampling methods?

Eugene used his students in his classes in the University, and it will be well representing the population, which are college students

We want to test whether the mean height of eighth graders is 66 inches. State the null and alternative hypotheses in terms of H0 and Ha

H0: The mean height of eighth graders equals 66 inches. HA: The mean height of eighth graders does not equal to 66 inches.

Descriptive statistics are used to describe a set of__________, including summary measures such as the mean.

Data

Calculate the mean and standard deviation and interpret your findings for the following set of data showing the diastolic blood pressure measurements for a sample of 9 individuals: 61, 63, 64, 69, 71, 77, 80, 81, and 95.

On average, the average distance of an individual data point is approximately 10.93 diastolic pressure points from the mean diastolic pressure of 73.44.

What type of variable is X = customer rating of restaurant service (excellent, good, fair, poor)?

Ordinal

What is the z-value for the 35% value?

P(Z>z)=35% P(Z>z)=0.35 P(Z<-0.39)=0.35 z=-0.39 Go to z-table and the z-value for 35% is -0.385

A numerical value computed from a population is called a(n)

Parameter

What do you call the whole subjects you are interested in for your research?

Population

The [ ] of a screening test (or symptom) is the probability that a subject does not have the disease, given that the subject has a negative screening test result (or does not have the symptom).

Predictive value negative

The [ ] of a screening test (or symptom) is the probability that a subject has the disease given that the subject has a positive screening test result (or has the symptom).

Predictive value positive

Which one of the following statistics is resistant to outliers?

Q1

Changing the μ of a distribution does what to the probability density function?

Shifts the distribution on the horizontal axis Shifts the distribution on the vertical axis Determines the spread of the distribution None of the options Correct answer is A

A statistical program output is shown below for a sample of n = 142 hourly employees on the variable X = job performance rating on a scale of 0.0 to 10.0. Use this output to answer the questions below the output. Descriptive Statistics: Rating Variable N Mean StDev Minimum Q1 Median Q3 Maximum Rating 142 7.91 1.56 3.40 7.10 8.30 9.00 10.00 Based on the output above, the most likely shape of the distribution of performance ratings is ____

Skewed left

A numerical value computed from a sample is called a

Statistic

Standard Deviation

The __________ is a measure of dispersion in original units. It is calculated as the square root of the variance.

Explain the empirical rule of the normal distribution.

The empirical rule of the normal distribution states that approximately 68% of the data falls within one standard deviation of the mean. 95% of the data falls within two standard deviations from the mean, and 99.7% of the data falls within three standard deviations of the mean.

Which of the following is NOT an assumption of the Binomial distribution?

The probability of success is equal to 0.5 in all trials

Which of the following best describes the meaning behind the area under the probability density function's smooth curve?

The shape and spread of the distribution Correct! The probability associated with the specified range of outcomes The range of possible outcomes The standard deviation of the distribution

Which of the following is the right explanation of the Normal distribution?

The standard deviation is 1, which we call z Correct! The shape is a bell-shape The shape is asymmetric The mean is always 0

Characteristics that a researcher label for a certain research topic is called _______ such as diastolic blood pressure, heart rate, and the heights of adult male, and so on.

Variable

Suppose the average length of stay in a chronic disease hospital for a certain type of patient is 60 days with a standard deviation of 15. It is reasonable to assume an approximately normal distribution of length of stay. Find the probability that a randomly selected patient from the group will have a length of stay between 30 and 60 days.

Z = (30-60) / (15) = -2. P ( z < -2) = 0.02275 = 0.0228 Z = (60-60) / (15) = 0. P ( z < 0 ) = 0.50000 = 0.5000 P ( -2 < z < 0 ) 0.5000 - 0.0228 = 0.4772

The [__________________] is the probability of rejecting a true null hypothesis.

level of significance

Normal distributions are centered on the value of the [ ].

mean

Of the three measures, which tends to reflect skewing the most, the mean, the mode, or the median?

mean

The distribution of results from a cholesterol test has a mean of 180 and a standard deviation of 20. A sample size of 40 is drawn randomly. a. Find the probability that the sum of the 40 values is greater than 7,500. b. Find the sum that is one standard deviation above the mean of the sums. c. Find the percentage of sums between 1.5 standard deviations below the mean of the sums and one standard deviation above the mean of the sums.

a) mean= 180 st.dv= 20 40 x 180 = 7200 sq rt of 40 x 20 = 126.49 7500-7200/126.49 = 2.37 P (z>2.37) = 1 - P(z < 2.37) z table for 2.37 = 0.9911 1 - 0.9911= 0.0089 b) 7200 + (1 x 126.49) = 7200 + 126.49 = 7326.49 c. Ex = N(40x180, 20 x sq rt of 40) P((7200-1.5(20 x sq rt of 40)) < Ex < (7200 + (20 x sq rt of 40)) = P(-1.5 < z <1) from z table, area between z scores -1.5 and 1 =0.77454

Different [______________] lead to modifications of confidence intervals.

assumptions

Marital status is best defined as which of the following types of variables?

categorical

The nature of the [_______] that form the basis of the testing procedures must be understood, since this determines the particular test to be employed.

data

stem and leaf plot

displays an ordered array of the data and are most effective with relatively small data sets. It provides information regarding the range of the data set, shows the location of the highest concentration of measurements and reveals the presence or absence of symmetry.

The standard deviation of any data set is the equivalent of the sample variance of the data set squared.

false

A researcher decides to take a random sample of individuals from the population and then track their sugar intake. Assuming the data collected on each individual will be classified into different levels of intake including little to no sugar intake, moderate sugar intake, and high sugar intake, which type of graph would allow the researcher to best represent the data they have collected?

histogram

A [____________] may be defined simply as a statement about one or more populations.

hypothesis

A researcher measures the amount of sugar in several cans of the same soda. The mean is 39.01 with a standard deviation of 0.5. The researcher randomly selects a sample of 100. a. Find the probability that the sum of the 100 values is less than 3,900. b. Find the sum with a z -score of -2.5. c. Find the probability that the sums will fall between the z -scores -2 and 1.

mean = 39.01, standard deviation = 0.5, n = 100 A) sample mean = 3900/100 = 39 0.5/ square root of 100 = 0.05 p ( z > (39 - 39.01 / 0.05) ) = -0.2 z score of -0.2 = 0.42074 B) -2.5 x 0.5/ square root of 100 + 39.01 = 38.885 38.885 x 100 = 3888.5 C) P ( -2 < z < 1) z score for - 2 = 0.02275 z score for 1 = 0.84134 P ( 0.02275 < z < 0.84134)

An unknown distribution has a mean of 80 and a standard deviation of 12. A sample size of 95 is drawn randomly from the population.a. Find the probability that the sum of the 95 values is greater than 7,650.b. Find the sum that is two standard deviations above the mean of the sums.

mean = 80, standard deviation = 12, n=95 a) 7650 / 95 = 80.526 P = 80.526 - 80 / (12 square root 95) = 0.4276 P(z > 0.4276) The probability that the sum of the 95 values is greater than 7650 is 0.66276 b) 95 x 80 + 2 square root 95 x 12 = 7,667.578 a) z table > 0.427 = 0.3347 P (sum of 95 > 7650) = 0.3347 b) sample size x mean + 2 x sq rt of sample size = 95 x 80 + 2 x sq rt of 95 = 7833.923

When the data are symmetrical, what is the typical relationship between the mean and median?

mean and median are close or the same

A statistical program output is shown below for a sample of n = 142 hourly employees on the variable X = job performance rating on a scale of 0.0 to 10.0. Use this output to answer the questions below the output. Descriptive Statistics: Rating Variable N Mean StDev Minimum Q1 Median Q3 Maximum Rating 142 7.91 1.56 3.40 7.10 8.30 9.00 10.00 Suppose the minimum value of 3.4 was changed to 2.0. Which one of the following statistics would not change?

median

Suppose we want to estimate the mean BMI for women in pregnancy at 20 weeks gestation. If we have a sample of 100 women and measure their BMI at 20 weeks gestation, what is the probability that the sample mean is within 1 unit of the true BMI if the standard deviation in BMI is taken to be 3.6?

n = 100, standard deviation = 3.6 P ( (28.5-1) - 28.5/3.6/Square root 100 < X < (28.5 + 1) - 28.5/3.6/square root 100 ) P ( -2.78 < X < 2.78) 0.9978 - 0.0027 0.9951 The probability that the sample mean is within 1 unit of the true BMI is 0.9951.

Arterial blood gas analyses performed on a sample of 15 physically active adult males yielded the following resting PaO2 values: 75; 80; 80; 74; 84; 78; 89; 72; 83; 76; 75; 87; 78; 79; 88 Compute the 95 percent confidence interval for the mean of the population.

n = 15, mean = 1198/15 = 79.866, standard deviation = 5.3031 95% confidence interval 1- 0.95 = 0.05 0.05 / 2 = 0.025 Degrees of freedom: 15-1 = 14 From the t-table = 2.145 5.3031 / square root of 15 = 1.369 79.86 plus 2.145 (5.3031/square root of 15) = 82.804 79.86 minus 2.145 (5.3031/square root of 15) = 76.930

A random sample of 16 emergency reports was selected from the files of an ambulance service. The mean time (computed from the sample data) required for ambulances to reach their destinations was 13 minutes. Assume that the population of times is normally distributed with a variance of 9. Can we conclude at the .05 level of significance that the population mean is greater than 10 minutes?

n = 16, mean = 13, variance = 9/ standard deviation = 3, alpha = 0.05, H0 = mean = 10 HA= mean greater than or equal to 10 Z = 13-10 / square root of 9/16 = 4 z = 4 4 in the Z table = 0.9997 or 1 1 - P(Z < 4) 1 -1 = 0.0003. P value is less than 0.05 significance level. Reject the null hypothesis that the population mean is 10. The population mean is actually greater than 10 minutes.

Following a week-long hospital supervisory training program, 16 assistant hospital administrators made a mean score of 74 on a test administered as part of the evaluation of the training program. The sample standard deviation was 12. Can it be concluded from these data that the population mean is greater than 70? Let α = .05. What assumptions are necessary?

n = 16, mean = 74, standard deviation = 12, alpha = 0.05 H0: μ less than or equal to 70 HA: μ > 70 test statistic t = 74 - 70/ 12/square root 16 4/3 =1.33 degrees of freedom = 15 critical t value: 1.7535 The test statistic, 1.33, is less than the critical t value, 1.7535. Thus, we fail to reject the null hypothesis. We cannot conclude from this data that the population mean is greater than 70. Assumptions necessary include: the data is from a random sample, and that the data is of normal distribution.

A study was made of a sample of 25 records of patients seen at a chronic disease hospital on an outpatient basis. The mean number of outpatient visits per patient was 4.8, and the sample standard deviation was 2. Can it be concluded from these data that the population mean is greater than four visits per patient? Let the probability of committing a type I error be .05. What assumptions are necessary?

n = 25, mean = 4.8, sample standard deviation = 2, alpha = 0.05 H0= μ = 4. HA= μ > 4 1) Test statistic t= 4.8-4/2square root 25 = 0.8/0.4 = 2.0 critical t-value: 1.7114 The critical t value is less than the test statistic, thus we reject the null hypothesis. Assumptions: the data is from a random sample, and the data is from a normal distribution. Professor's answer: H0 ≤ 4 Ha > 4

Suppose it is known that the IQ scores of a certain population of adults are approximately normally distributed with a standard deviation of 15. A simple random sample of 25 adults drawn from this population had a mean IQ score of 105. On the basis of these data can we conclude that the mean IQ score for the population is not 100? Let the probability of committing a type I error be .05.

n = 25, sample standard deviation = 15, sample mean = 105, alpha = 0.05 H0 = u = 100 HA= u does not equal 100. test statistic: Z = 105 - 100 / 15/ Square root of 25 z = 1.67 Two-tailed test, so alpha 0.05 = 1.96. Since 1.67 is less than 1.96, we fail to reject the null hypothesis. The mean IQ score for the population is not 100.

Nine laboratory animals were infected with a certain bacterium and then immunosuppressed. The mean number of organisms later recovered from tissue specimens was 6.5 (coded data) with a standard deviation of .6. Can one conclude from these data that the population mean is greater than 6? Let α = .05. What assumptions are necessary?

n = 9, mean = 6.5, standard deviation = 0.6, alpha = 0.05 H0= μ = 6, HA= μ > 6 test statistic: t = 6.5 - 6.0/ 0.6/square root of 9 0.5 / 0.2 = 2.5 looking at t table, the t value of 2.5 lies between 2.306 and 2.896. The p values of this are 0.025 and 0.01 for 8 degrees of freedom. Since 0.05 is greater than 0.025 and 0.01, the null hypothesis is rejected. Assumptions that are necessary include data from a random sample and data that is normally distributed.

A false negative results when a test indicates a [ ] status when the true status is positive

negative

How many different Standard Normal distributions are there? [ ]

one

The total area under a Normal curve sums to exactly [ ]

one

What is the standard deviation of the Standard Normal distribution? [ ]

one

A study is being conducted to determine the prevalence of lung cancer in a population and the researchers want to note the frequency of each particular stage of cancer among the members of the population who have lung cancer; in this case the stage of cancer is best defined as which of the following types of variables?

ordinal variable

Inferential statistics are used to infer or predict information about a blank based blank on data collected from a .

population and sample

A false positive results when a test indicates a [ ] status when the true status is negative.

positive

All possible values that the test statistic can assume are points on the horizontal axis of the graph of the distribution of the test statistic and are divided into two groups; one group constitutes what is known as the and the other group makes up the .

rejected region / non-rejected region

The [____________] is the conjecture or supposition that motivates the research.

research hypothesis

The [ ] of a test (or symptom) is the probability of a positive test result (or presence of the symptom) given the presence of the disease.

sensitivity

The [ ] of a test (or symptom) is the probability of a negative test result (or absence of the symptom) given the absence of the disease.

specificity

Normal distribution has two parameters: name them:

standard deviation and mean

[________________] are hypotheses that are stated in such a way that they may be evaluated by appropriate statistical techniques.

statistical hypothesis

Which of the following is NOT explaining about the normal distribution?

t has two parameters, number of trials and success rate. Its range is - ∞ through ∞ It is a continuous variable. It is a symmetric shape. A is the answer

The [______________] is some statistic that may be computed from the data of the sample.

test statistic