For a school, the principal expected that a student takes 45 minutes to travel to school on average. Suppose the time taken to travel to school by a student follows a normal distribution. 100 students are randomly selected from the school for interview and they took on average 44 minutes to travel to school with a standard deviation of 15 minutes. 6 of the selected students had lateness record in the last month.

1. Construct a 95% confidence interval for the true mean time taken to travel to school by a student. Correct the final answers to 2decimal places.
2. How many additional sample is needed to ensure the estimate of μ is accurate within 0.8 minutes with a confidence of 94% and σ is known to be 10 minutes.
3. By using critical value approach to perform a hypothesis testing for the principal’s claim that the true mean time taken to travel to school by student was 45minutes at level of significant α= 0.05. (if σ is unknown) Correct the final answers to 2 decimal places.
4. Construct a 92.5% confidence interval for the true proportion (p) of students who had lateness record in the last month. Correct the final answers to 4decimal places.
5. How large a sample is required to ensure the estimate of p is accurate within 2% with a confidence level of 97%?
6. By using critical value approach to perform a hypothesis testing for the principal’s claim that under (10 of all students had lateness record in the last month at level of significant α = 0.01. Correct the test statistics to 2 decimal places.

a. What statistical procedure will solve this problem? A sociologist is interested in determining if the number of young adults in a city is related to the number of burglaries that occurred in the adult shops in the same city last year.

Group of answer choices

dependent t

Pearson r correlations

independent t

b. What statistical procedure will solve this problem? Just before retiring from teaching college, a professor wanted to know if his former students’ final grade scores (which he diligently kept for over 30 years) were related to the subsequent income earned by the same students (which he also researched and recorded over the same period).

Group of answer choices

dependent t

Pearson r correlations

independent t

c. Identify the DEPENDENT VARIABLE in this problem: Pomona College Researchers conducted a study to determine if chewing gum benefits grade scores of students that suffer with ADHD.

Group of answer choices

students with ADHD

chewing gum

focus

grade scores

d. Identify the DEPENDENT VARIABLE in this problem: Does lung mass change in mice if they are placed at a high altitude?

Group of answer choices

lung mass

high altitude

mice

e. Identify the INDEPENDENT VARIABLE in this problem: Does owning a pet significantly reduce depression.

Group of answer choices

depression

affect

owning a pet

In a survey of 3470 adults aged 57 through 85​ years, it was found that 82.3​% of them used at least one prescription medication. Complete parts​ (a) through​ (c) below. a. How many of the 3470 subjects used at least one prescription​ medication? nothing ​(Round to the nearest integer as​ needed.) b. Construct a​ 90% confidence interval estimate of the percentage of adults aged 57 through 85 years who use at least one prescription medication. nothing​%less thanpless than nothing​% ​(Round to one decimal place as​ needed.) c. What do the results tell us about the proportion of college students who use at least one prescription​ medication? A. The results tell us that there is a​ 90% probability that the true proportion of college students who use at least one prescription medication is in the interval found in part​ (b). B. The results tell us nothing about the proportion of college students who use at least one prescription medication. C. The results tell us​ that, with​ 90% confidence, the probability that a college student uses at least one prescription medication is in the interval found in part​ (b). D. The results tell us​ that, with​ 90% confidence, the true proportion of college students who use at least one prescription medication is in the interval found in part​ (b).

A study was designed to compare the attitudes of two groups of nursing students towards computers. Group 1 had previously taken a statistical methods course that involved significant computer interaction. Group 2 had taken a statistic methods course that did not use computers. The students' attitudes were measured by administering the Computer Anxiety Rating Scale (CARS). A random sample of 11 nursing students from Group 1 resulted in a mean score of 56.1 with a standard deviation of 5.5 A random sample of 16 nursing students from Group 2 resulted in a mean score of 67.7 with a standard deviation of 7.3 Can you conclude that the mean score for Group 1 is significantly lower than the mean score for Group 2? Let μ1μ1 represent the mean score for Group 1 and μ2 represent the mean score for Group 2. Use a significance level of α=0.05 for the test. Assume that the population variances are equal and that the two populations are normally distributed.

Step 1 of 4: State the null and alternative hypotheses for the test.

Step 2 of 4: Compute the value of the t test statistic. Round your answer to three decimal places.

Step 3 of 4: Determine the decision rule for rejecting the null hypothesis H0. Round your answer to three decimal places.

Step 4 of 4: State the test's conclusion.

The Gourmand Cooking School runs short cooking courses at its small campus. Management has identified two cost drivers it uses in its budgeting and performance reports—the number of courses and the total number of students. For example, the school might run two courses in a month and have a total of 64 students enrolled in those two courses. Data concerning the company’s cost formulas appear below:

For example, administrative expenses should be $4,000 per month plus $44 per course plus $5 per student. The company’s sales should average $870 per student.

The company planned to run four courses with a total of 64 students; however, it actually ran four courses with a total of only 60 students. The actual operating results for September appear below:

Required:

1. Prepare the company’s planning budget for September.

2. Prepare the company’s flexible budget for September.

3. Calculate the revenue and spending variances for September.

The MCAT, Medical College Admissions Test, must be taken by students wishing to gain entrance to most Medical Programs in Canada and the United States. There are four sections: Physical Sciences (PS), Verbal Reasoning (VR), Biological Sciences (BS), and Writing Sample (WS). The First three are each scored on a scale from 1 to 15, with a mean of 8 and a standard deviation of 2.

(a)Determine the percentile ranks for the following students:

(b)If a Medical School sets a requirement that students who will be considered must have a score of 10 on the Physical Science, 12on the Verbal Reasoning and 11 on the Biological Sciences Sections, what proportion of the population will qualify for each portion of the test?

3.The ACT, American Colleges Test, is comparable to the SAT ReasoningTest. The ACT is well modeled by a normal distribution, X~ N(21, 52),whereas the SAT Reasoning Test is modeled by S~N(1500, 3002). A college will accept either the SAT or the ACT for admissions. A college admissions assistant needs to compare the following two students to each other though they wrote different tests. How would the admissions assistant do this? Which student seemed to score better?

Matthew Chant ACT Score = 27

Joseph BellSAT Score = 2050

A study was designed to compare the attitudes of two groups of nursing students towards computers. Group 1 had previously taken a statistical methods course that involved significant computer interaction. Group 2 had taken a statistic methods course that did not use computers. The students' attitudes were measured by administering the Computer Anxiety Rating Scale (CARS). A random sample of 12 nursing students from Group 1 resulted in a mean score of 54.7 with a standard deviation of 6. A random sample of 15 nursing students from Group 2 resulted in a mean score of 66.6 with a standard deviation of 3. Can you conclude that the mean score for Group 1 is significantly lower than the mean score for Group 2? Let μ1 represent the mean score for Group 1 and μ2 represent the mean score for Group 2. Use a significance level of α=0.1 for the test. Assume that the population variances are equal and that the two populations are normally distributed. Step 1 : State the null and alternative hypotheses for the test. Step 2 : Compute the value of the t test statistic. Round your answer to three decimal places Step 3 : Determine the decision rule for rejecting the null hypothesis H0. Round your answer to three decimal places. Reject or fail to reject your hypothesis?

2. The MCAT, Medical College Admissions Test, must be taken by students wishing to gain entrance to most Medical Programs in Canada and the United States. There are four sections: Physical Sciences (PS), Verbal Reasoning (VR), Biological Sciences (BS), and Writing Sample (WS). The first three are each scored on a scale from 1 to 15, with a mean of 8 and a standard deviation of 2.

(a) Determine the percentile ranks for the following students:

(b) If a Medical School sets a requirement that students who will be considered must have a score of 10 on the Physical Science, 12 on the Verbal Reasoning and 11 on the Biological Sciences sections, what proportion of the population will qualify for each portion of the test?

c) The ACT, American Colleges Test, is comparable to the SAT Reasoning Test. The ACT is well modeled by a normal distribution, X~ N(21, 52 ), whereas the SAT Reasoning Test is modeled by S~N(1500, 3002 ). A college will accept either the SAT or the ACT for admissions. A college admissions assistant needs to compare the following two students to each other though they wrote different tests. How would the admissions assistant do this? Which student seemed to score better? Matthew Chant ACT Score = 27 Joseph Bell SAT Score = 2050

A study was designed to compare the attitudes of two groups of nursing students towards computers. Group 1 had previously taken a statistical methods course that involved significant computer interaction. Group 2 had taken a statistic methods course that did not use computers. The students' attitudes were measured by administering the Computer Anxiety Rating Scale (CARS). A random sample of 10 nursing students from Group 1 resulted in a mean score of 65.8 with a standard deviation of 5.1. A random sample of 16 nursing students from Group 2 resulted in a mean score of 70.4 with a standard deviation of 7.6. Can you conclude that the mean score for Group 1 is significantly lower than the mean score for Group 2? Let μ1 represent the mean score for Group 1 and μ2 represent the mean score for Group 2. Use a significance level of α=0.05 for the test. Assume that the population variances are equal and that the two populations are normally distributed.

Step 1 of 4: State the null and alternative hypotheses for the test.

Step 2 of 4: Compute the value of the t test statistic. Round your answer to three decimal places.

Step 3 of 4: Determine the decision rule for rejecting the null hypothesis H0. Round your answer to three decimal places.

Step 4 of 4: State the test's conclusion.

Drew wondered if the average age of students in AP Statistics classes in his high school is under 18. He randomly selected 10 AP Statistics students in his school and the data set is given below. Using the information from a previous survey, he has concluded that the population standard deviation for the age of AP Statistics students in his school is 1.23. Assume that the ages of students in AP Statistics classes in this high school are normally distributed.

A calculator was used to determine the p-value for this hypothesis test. The p-value was 0.005. If the level of significance was 0.01, interpret the results.

The p-value, 0.005, is less than the level of significance, 0.01. Reject the null hypothesis that the mean age is 18. There is sufficient evidence to conclude that the mean age is under 18.

The p-value, 0.005, is less than the level of significance, 0.01. Reject the null hypothesis that the mean age is 18. There is insufficient evidence to conclude that the mean age is under 18.

The p-value, 0.005, is less than the level of significance, 0.01. Do not reject the null hypothesis that the mean age is 18. There is insufficient evidence to conclude that the mean age is under 18.

The p-value, 0.005, is less than the level of significance, 0.01. Do not reject the null hypothesis that the mean age is 18. There is sufficient evidence to conclude that the mean age is under 18.