What type of study is this research and why?
Identify the variables.
Evaluation the internal validity.
What type of study is this research and why?
Identify the variables.
Evaluation the internal validity
What type of study is this research and why?
Identify the variables.
Evaluation the internal validity.
What type of study is this research and why?
Identify the variables.
Evaluation the internal validity.
In: Statistics and Probability
A class survey in a large class for first-year college students asked, "About how many minutes do you study on a typical weeknight?" The mean response of the 261 students was x¯¯¯x¯ = 130 minutes. Suppose that we know that the studey time follows a Normal distribution with standard deviation σσ = 65 minutes in the population of all first-year students at this university. Regard these students as an SRS from the population of all first-year students at this university. Does the study give good evidence that students claim to study more than 2 hours per night on the average?
(a) State null and alternative hypotheses in terms of the mean
study time in minutes for the population.
(b) What is the value of the test statistic z?
(c) Can you conclude that students do claim to study more than two
hours per weeknight on the average?
(a) H0: Ha: (Type in "mu" as the
substitute for μμ and "!=" for ≠≠.)
(b) z:
(c) Conclusion: (Answer with "Yes/Y" or "No/N".)
A study is conducted to determine if a newly designed text book is more helpful to learning the material than the old edition. The mean score on the final exam for a course using the old edition is 75. Ten randomly selected people who used the new text take the final exam. Their scores are shown in the table below.
| Person | A | B | C | D | E | F | G | H | I | J |
| Test Score | 85 | 78 | 70 | 88 | 67 | 96 | 74 | 90 | 81 | 93 |
Use a 0.010.01 significance level to test the claim that people do
better with the new edition. Assume the standard deviation is 10.3.
(Note: You may wish to use statistical software.)
(a) What kind of test should be used?
A. Two-Tailed
B. One-Tailed
C. It does not matter.
(b) The test statistic is (rounded to 2 decimals).
(c) The P-value is
(d) Is there sufficient evidence to support the claim that
people do better than 75 on this exam?
A. Yes
B. No
(e) Construct a 9999% confidence interval for the mean score for
students using the new text.
<μ<<μ<
In: Statistics and Probability
The three charges are at the three vertices of an equilateral triangle ?( all angles are 60degrees)
q 1 = + 10.0 µC
q 2 = - 5 .0 nC
q 3 = + 8 .0 nC
Equilateral side of the triangle = 0.05 m.
A. Draw forces acting on q 1 by q 2and q 3
B. find the components on X and Y axes .
C. Use Pythagorean theorem to find the resultant .
D. Use tangent to find the direction ( angle ) the resultant makes with horizontal.
q 3 q 1 q 2
2-A proton is released from rest between two conducting plates. The distance between the two plates is 6.0 cm how fast will the electron be moving reaches to the second plate. The potential difference between the two plates is 10 KV.
= 1.60 × 10-19C, m proton= 1.67 × 10-27kg)
Show formulas, substitution, and calculation with units.
-What are the currents in each resistor?
Voltage across the battery V = 12 volts
R1= 1?, R2= 2?, R3= 3?, R4= 4?, R5= 5?, R6= 6?,
4. In an RCseries circuit, V = 24.0 V, resistance R= 47 K?, and capacitance C= 2200 ?F.
Calculate the time constant.
Define time constant in a complete sentence.
C) How long does it take for the capacitor to reach to its 63% of its maximum charge?
d) How long does it take for the capacitor to reach to its 75% of it maximum charge?
e) How much charge can be stored in the capacitor after 50 seconds?
5. Find the electric potential of the following configuration at point P. The two short sides of the triangle are the same size = 23 cm.
Q1 = - 5600 nC
Q2 = - 2200 nC
Q1 Q2 p
F= K Q1Q2/ r2
E = K Q/ r2
V= K Q/ r
U or PE = Q V
KE = ½ m v2
R = R1+ R2+ ….
1/R = 1/R1+ 1/R2+….
FY F FX
FX= F Cos?
FY= F Sin?
F2= FX 2 + FY 2
Q = Qm ( 1/ e- t/ RC )
Q = C V
In: Physics
Suppose we need to construct a tin can with a fixed volume V cm3 in the shape of a cylinder with radius r cm and height h cm. (Here V should be regarded as a constant. In some sense, your answers should be independent of the exact value of V .) The can is made from 3 pieces of metal: a rectangle for the side and two circles for the top and bottom. Suppose that these must be cut out of a rectangular sheet of metal. Our goal is to find the values of r and h, and the dimensions of this rectangular sheet that minimize its area.
Draw a picture of how the rectangle and two circles could be cut out of a larger rectangle. There are multiple ways to do this (I can think of at least 3). Draw as many as you can, solve the problems below for each arrangement and then compare your answers.
Label the sides of the rectangle in terms of r and h. Express the rectangle’s area in terms of r and h. Also, note whether there are any assumptions about r and h that you need to make in order for your picture to make sense. (For example, if you draw a circle with diameter 2r inside of a rectangle with side l, then you must have 2r ≤ l.)
Use the fact that the can’s volume is V = πr2h to express h in terms of r, and write the rectangle’s area as a function of r. (Or else, you may alternatively solve for r and write the area as a function of h.)
Find the value of r (or h) that minimizes the rectangle’s area. What is the correspond- ing value of h (or r), and the dimensions of the rectangle? Your answers will most likely be in terms of V , but the ratio h/r might be a number. What is the minimum area of the rectangle in terms of V ?
As mentioned above, you should complete (1)-(4) for as many different arrangements as you can think of. (The math for some might be very simple.) Then compare your answers to find the best way of arranging the 2 circles and rectangle inside the larger rectangle, and the minimum possible area of the rectangle.
What if you need to make 2 (or more) cans in the same way. Can you find an arrangement of all the necessary pieces inside a single rectangle that is even more efficient?
In: Physics
package SOLUTION;
import java.util.Collection;
import java.util.Set;
public interface MapInterface<K, V> {
/**
* If the given key is not already in the map, adds
the
* key-value pair to the map. Otherwise, updates the
old
* value of the existing key to the specified
value.
* @param key the key
* @param value the value to be stored in the map with
the key
* @return null if the key was not already in the map,
or
* the old value associated with the key if the key was
already in the map
*/
public V put(K key, V value);
/**
* Gets the value from the map that is associated with
the given key
* @param key the key
* @return the value associated with the key, or null
if the key is
* not in the map
*/
public V get(K key);
/**
* Removes from the key-value pair associated with the
specified key
* @param key the key
* @return the value associated with the key, or null
if the key is
* not in the map
*/
public V remove(K key);
/**
* Returns whether the map contains the key-value pair
associated with
* the specified key
* @param key the key
* @return true if the map contains a key-value pair
with the specified
* key, and false otherwise
*/
public boolean containsKey(K key);
/**
* Returns whether the map contains no elements
* @return true if the map contains no key-value pairs,
and false otherwise
*/
public boolean isEmpty();
/**
* Removes all elements from the map
*/
public void clear();
/**
* Gets the number of key-value pairs in the map
* @return the number of key-value pairs in the
map
*/
public int size();
/**
* Gets a set of all keys in the map
* @return
*/
public Set<K> keySet();
/**
* Gets a set of all values in the map
* @return
*/
public Collection<V> values();
}
package SOLUTION;
public class Tester {
public static void main(String[] args) {
ArrayListMap<String, Integer>
alm = new ArrayListMap<String, Integer>();
System.out.println(alm.put("Cindy",
1));
System.out.println(alm.put("Nina",
2));
System.out.println(alm.put("Morgan", 3));
System.out.println(alm.put("Michael", 4));
System.out.println(alm.size());
System.out.println(alm.containsKey("Morgan"));
System.out.println(alm.containsKey("Jack"));
System.out.println(alm.get("Michael"));
System.out.println(alm.values());
System.out.println(alm.keySet());
System.out.println(alm.put("Michael", 6));
System.out.println(alm.get("Michael"));
System.out.println(alm.remove("Michael"));
System.out.println(alm.remove("Morgan"));
System.out.println(alm.remove("Nina"));
System.out.println(alm.remove("Cindy"));
System.out.println(alm.size());
}
}
In: Computer Science
Student Debt – Vermont: The average student loan debt of a U.S. college student at the end of 4 years of college is estimated to be about $22,500. You take a random sample of 146 college students in the state of Vermont and find the mean debt is $23,500 with a standard deviation of $2,600. We want to construct a 90% confidence interval for the mean debt for all Vermont college students.
(a) What is the point estimate for the mean debt of all Vermont
college students?
$
(b) What is the critical value of t (denoted
tα/2) for a 90% confidence interval?
Use the value from the table or, if using software, round
to 3 decimal places.
tα/2 =
(c) What is the margin of error (E) for a 90% confidence
interval? Round your answer to the nearest whole
dollar.
E = $
(d) Construct the 90% confidence interval for the mean debt of all
Vermont college students. Round your answers to the nearest
whole dollar.
< μ <
(e) Based on your answer to (d), are you 90% confident that the
mean debt of all Vermont college students is greater than the
quoted national average of $22,500 and why?
No, because $22,500 is below the lower limit of the confidence interval for Vermont students.No, because $22,500 is above the lower limit of the confidence interval for Vermont students. Yes, because $22,500 is below the lower limit of the confidence interval for Vermont students.Yes, because $22,500 is above the lower limit of the confidence interval for Vermont students.
(f) We are never told whether or not the parent population is
normally distributed. Why could we use the above method to find the
confidence interval?
Because the margin of error is positive.Because the margin of error is less than 30. Because the sample size is less than 100.Because the sample size is greater than 30.
In: Statistics and Probability
Using diaries for many weeks, a study on the lifestyles of visually impaired students was conducted. The students kept track of many lifestyle variables including how many hours of sleep obtained on a typical day. Researchers found that visually impaired students averaged 8.82 hours of sleep, with a standard deviation of 2.12 hours. Assume that the number of hours of sleep for these visually impaired students is normally distributed.
(a) What is the probability that a visually impaired student gets at most 6.2 hours of sleep? Express your answer as a percent rounded to 2 decimal places. e.g. 1.23% Do not include the % symbol in your answer.
Answer: ____%
(b) What is the probability that a visually impaired student gets between 6.6 and 10.28 hours of sleep? Express your answer as a percent rounded to 2 decimal places. e.g. 1.23% Do not include the % symbol in your answer.
Answer: _____%
(c) What is the probability that a visually impaired student gets at least 7.3 hours of sleep? Express your answer as a percent rounded to 2 decimal places. e.g. 1.23% Do not include the % symbol in your answer.
Answer: ______%
(d) What is the sleep time that cuts off the top 38.4% of sleep hours? Round your answer to 2 decimal places.
Answer: _____hours
(e) If 100 visually impaired students were studied, how many students would you expect to have sleep times of more than 10.28 hours? Round to the nearest whole number.
Answer: _____students
(f) A school district wants to give additional assistance to visually impaired students with sleep times at the first quartile and lower. What would be the maximum sleep time to be recommended for additional assistance? Round your answer to 2 decimal places.
Answer: ______hours
In: Statistics and Probability
An organization monitors many aspects of elementary and secondary education nationwide. Their 1995 numbers are often used as a baseline to assess changes. In 1995, 42 % of students had not been absent from school even once during the previous school year. In the 1999 survey, responses from 7146 randomly selected students showed that this figure had slipped to 41 %. Officials would note any change in the rate of student attendance. Answer the questions below.
(a) Write appropriate hypotheses.
Upper H 0 : The percentage of students in 1999 with perfect attendance the previous school year (1)_______
Upper H Subscript Upper A Baseline : The percentage of students in 1999 with perfect attendance the previous school year (2)_________
(b) Check the necessary assumptions and conditions.
The independence assumption is (3)___________
The randomization condition is (4) __________
The 10% condition is (5)____________
The success/failure condition is (6) ______________
(c) Perform the test and find the P-value.
P-value equals _____________ (Round to three decimal places as needed.)
(d) State your conclusion. Consider probabilities less than 0.05 to be suitably unlikely.
A. We fail to reject the null hypothesis. There is not sufficient evidence to suggest that the percentage of students with perfect attendance in the previous school year has changed.
B. We fail to reject the null hypothesis. There is sufficient evidence to suggest that the percentage of students with perfect attendance in the previous school year has changed.
C. We can reject the null hypothesis. There is sufficient evidence to suggest that the percentage of students with perfect attendance in the previous school year has changed.
(1) is greater than 42%.
is different from 42%.
is less than 42%.
is equal to 42%.
(2) is less than 42%.
is different from 42%.
is greater than 42%.
is equal to 42%.
(3) satisfied. not satisfied.
(4) satisfied. not satisfied.
(5) not satisfied. satisfied.
(6) satisfied. not satisfied.
In: Statistics and Probability
Some statistics students estimated that the amount of change daytime statistics students carry is exponentially distributed with a mean of $0.72. Suppose that we randomly pick 25 daytime statistics students.
a. In words, define the random variable X.
(from these) -the number of coins that a daytime statistics student carries--the total amount of money that a daytime statistics student carries---the amount of change that a daytime statistics student carries---the number of daytime statistics students who carry change
b. Give the distribution of X.
c. In words, define the random variable x-bar
(from these)-the average number of daytime statistics students----the average number of daytime statistics students who carry change---the average amount of change a daytime statistics student carries---the average amount of money a daytime statistics student carries
d. Give the distribution of X bar (Round your standard deviation to three decimal places.)
e. Find the probability that an individual had between $0.72 and $0.90. (Round your answer to four decimal places.)
f. Find the probability that the average of the 25 students was between $0.72 and $0.90. (Round your answer to four decimal places.)
g. Explain why there is a difference between parts (e) and (f).
(from these)- The graph of sample means becomes more normal as the sample increases; therefore, in part (f), the graph is approximately normal and in part (e), the graph is exponential.-----The graph of the amount of change becomes more normal as the sample increases; therefore, in part (e), the graph is approximately normal and in part (f), the graph is exponential.----- There is always a better chance for 25 people to have between $0.72 and $0.90 than for 1 person to have that amount.-----There is always a better chance for one person to have between $0.72 and $0.90 than for 25 people to each have that amount.
In: Statistics and Probability
Independent samples t-test: An experimenter is interested in how the “foot-in-the-door” tactic could increase compliance in college students. The “foot-in-the-door” tactic involves asking first for a small request to butter someone up for a larger (originally intended) request. Typically, the “foot-in-the-door” tactic increases compliance because people do not like to appear inconsistent and since they originally agreed to the small request, they feel compelled to also agree with the large request.
Our experimenter was seeking to increase the number of students who volunteer for research in the department of psychological sciences. One group of students were subject to the “foot-in- the-door” tactic and these students were first asked to wear a sticker promoting research participation (small request) around campus. Later, these same students were asked to participate in a research project. The second group of students were simply asked to participate in research with no initial, small request.
This study was conducted across multiple semesters, and the value below represents how many students agreed to participate in research each semester.
“foot-in-the-door” (FITD): 11 8 10 6 5
Large request only (Large): 6 6 9 10 8
Now we’ll use this information for hypothesis testing....
What are your degrees of freedom for this independent samples t-test? 2pts.
What is your critical t-value (t-crit) for this hypothesis test? 2pts.
Next we calculate the t-value for our sample to compare to this critical value (see table below for reminders)......
What is the pooled variance for our sample? 4 pts. Pooled Variance: ??1+ ??2 =
?? 1+ ?? 2
What is the standard error of the mean difference (??1−?2)for our sample? 4 pts.
What is the value for t-observed? 4 pts.
Is there a statistically significant difference between these groups? 3pts.
In: Statistics and Probability