Mr. Cooper recently gave a test to his 50 student history class. The scores were normally distributed with a mean of 75 and a standard deviation of 7.

1) What percentage of students scored higher than​ 85?​% (Round to two decimal​ places)

2) What percentage of students scored between 73 and​ 80?% ​(round to two decimal​ places)

3) 75% of the students scored higher than what​ score? ​(Round to two decimal​ places)

​4) 30% of students scored lower than what​ score? (Round to two decimal​ places)

5) Approximately how many students got a C​ (70) or higher on the​ exam?  nothing

6) Approximately how many students got an A​ (90+) on the​ exam?

A student researcher compares the heights of American students and non-American students from the student body of a certain college in order to estimate the difference in their mean heights. A random sample of 12 American students had a mean height of 67.7 inches with a standard deviation of 3.06 inches. A random sample of 17 non-American students had a mean height of 64.7 inches with a standard deviation of 1.97inches. Determine the 90% confidence interval for the true mean difference between the mean height of the American students and the mean height of the non-American students. Assume that the population variances are equal and that the two populations are normally distributed.

Step 3 of 3 :  

Construct the 90%90% confidence interval. Round your answers to two decimal places.

4. The National Assessment of Education Progress (NAEP) tested a sample of students who had used a computer in their mathematics classes and another sample of students who had not used a computer. The sample mean score for students using the computer was 309 with a sample standard deviation 29. For students not using computer, the sample mean was 303 and sample standard deviation was 32. Assume there were 60 students in the computer sample and 40 students in the sample that hadn’t used a computer. Can you conclude that the population mean scores differ? Use a α=0.05 significance level.

a. Calculate the Margin of error for 99% confidence level.

b. Construct a 99% confidence interval for the difference in means

c. Do the Hypothesis test.

A student researcher compares the heights of American students and non-American students from the student body of a certain college in order to estimate the difference in their mean heights. A random sample of 17 American students had a mean height of 70 inches with a standard deviation of 2.87 inches. A random sample of 12 non-American students had a mean height of 65.1 inches with a standard deviation of 2.68 inches. Determine the 90% confidence interval for the true mean difference between the mean height of the American students and the mean height of the non-American students. Assume that the population variances are equal and that the two populations are normally distributed. Step 1 of 3 : Find the point estimate that should be used in constructing the confidence interval.

5. A national survey conducted in 2005 on Canadian undergraduate students with a questioner asking them whether they do part-time jobs. 1250 students participated in the survey and 802 students said they do part-time jobs. In 2020, researcher claims that there is an increase in undergraduate students doing part-time jobs due to increase in tuition fees. In January 2020, he found 963 students out of randomly selected 1420 students do part-time jobs. Do a hypothesis test at 7% significance level to test the researcher’s claim. Answer the following to do the test:

   1. State null and alternative hypotheses.

   2. State your decision rule.

   3. Calculate the test statistic.

   4. State your conclusion.

   5. Find the p-value of the test.

Grades on a standardized test are known to have a mean of 500 for students in the US. The test is administered to 600 randomly selected students in Florida. In this subsample, the mean is 508, and the standard deviation is 75

i. Construct a 95% confidence interval for the average test score for students in Florida.

ii. Is there statistically significant evidence that students in Florida perform differently from other students in the US?

iii. Another 500 students are selected at random from Florida. They are given a 3 hour preparation course before the test is administered. Their average test score is 514, with a standard deviation of 15. Construct a 95% confidence interval. Is there statistically significant evidence that the preparation course helped? What conditions must be met in order for the results to have a causal interpretation?

Harry Potter books have become popular with children and adults alike. A recent survey conducted in London revealed that 80% high school students have read the first Harry Potter book. A random sample of 7 London high school students is taken, and the number of students who have read the first Harry Potter book is recorded.

a) Define the random variable of interest and give its distribution, including the values of all the parameters.                          

X = (Click to select)the sample meanSample size of London high school students who participated in the surveyProbability that a London high school student has read the first Harry Potter bookNumber of London high school students who have read the first Harry Potter book

X ~ (Click to select)BinomialNormalContinuousUniform(n=, p=)

Round answers for parts (b) and (c) to two decimal places. Round answers for parts (d) through (h) to four decimal places.

b) What is the expected number of randomly selected students who have read the first Harry Potter book?   

c) What is the variance of number of students who have read the first Harry Potter book?   

d) What is the probability that exactly two of the randomly selected students has read the first Harry Potter book?

e) What is the probability that at least two of the randomly selected students have read the first Harry Potter book?

f) What is the probability that no more than five of the randomly selected students have read the first Harry Potter book?

g) What is the probability that between two and seven (inclusive) of the randomly selected students have read the first Harry Potter book?

h) What is the probability that more than seven of the randomly selected students have read the first Harry Potter book?

When an introductory statistics course has been taught live, the average final numerical grade has historically been 71.6% with a standard deviation of 8.8%. Forty students were selected at random to take a new online version of the same course, and the average final numerical grade for these students was found to be 69.2%. Assume that the grades for the online students are normally distributed and that the population standard deviation is still 8.8%. In parts (a) and (b), use the p-value method to conduct an appropriate hypothesis test at the 5% level of significance.

(a) Is there sufficient evidence to indicate that students taking the online version of the course score lower, on average, than students taking the traditional live version of the course?

(b) Is there sufficient evidence to indicate that the average score for students taking the online version of the course differs from students taking the traditional live version of the course? (HINT: You do not need to recalculate the value of the test statistic.)

According to the National Institute on Alcohol Abuse and Alcoholism, 19% of college students aged 18 to 24 abuse alcohol.1

Describe the population proportion of interest in words. What value are we assuming for this proportion?

What is the sampling distribution for the sample proportion of college students aged 18 to 24 that abuse alcohol from a random sample of size 100 from this population? Make sure to explain why the appropriate conditions are met or not met as part of your answer.

In a random sample of 100 college students aged 18 to 24, 22 abused alcohol. Compute the statistic of interest.

What is the probability of obtaining a sample of 100 college students aged 18 to 24 that has less than 22 students that abuse alcohol?

What is the probability of obtaining a sample of 100 college students aged 18 to 24 that has a sample proportion of students that binge drink more than 0.25?

1. Researchers obtained a sample of 36 college students who all have the same history instructor this semester. Half of the students were shown a 2-min video that claimed the purpose of education was to help students “learn how to learn” so that they can enjoy a lifetime of learning after college. other half of students were shown a 2-min video that claimed the purpose of education was to teach facts to students. After watching the 2-min videos, the students were asked to rate their history instructor using a 10-point scale, 1 = very bad teacher to 10 = very good teacher. The mean and standard deviations for each group of 18 students are provided below. Use the provided information to answer the next four questions. Use an α of .05, two tailed.

“learn to learn” Group 1: M₁ = 5.9, SD₁ = 1.8, n₁ = 18

“learn facts” Group 2: M₂ = 7.2, SD₂ = 1.7, n₂ = 18

Compute the effect size of this study.

2. Researchers obtained a sample of 36 college students who all have the same history instructor this semester. Half of the students were shown a 2-min video that claimed the purpose of education was to help students “learn how to learn” so that they can enjoy a lifetime of learning after college. The other half of students were shown a 2-min video that claimed the purpose of education was to teach facts to students. After watching the 2-min videos, the students were asked to rate their history instructor using a 10-point scale, 1 = very bad teacher to 10 = very good teacher. The mean and standard deviations for each group of 18 students are provided below. Use the provided information to answer the next four questions. Use an α of .05, two tailed.

“learn to learn” Group 1: M₁ = 5.9, SD₁ = 1.8, n₁ = 18

“learn facts” Group 2: M₂ = 7.2, SD2 = 1.7, n₂ = 18

How large is the effect size?

3. Researchers obtained a sample of 36 college students who all have the same history instructor this semester. Half of the students were shown a 2-min video that claimed the purpose of education was to help students “learn how to learn” so that they can enjoy a lifetime of learning after college. The other half of students were shown a 2-min video that claimed the purpose of education was to teach facts to students. After watching the 2-min videos, the students were asked to rate their history instructor using a 10-point scale, 1 = very bad teacher to 10 = very good teacher. The mean and standard deviations for each group of 18 students are provided below. Use the provided information to answer the next four questions. Use an α of .05, two tailed.

“learn to learn” Group 1: M₁ = 5.9, SD₁ = 1.8, n₁ = 18

“learn facts” Group 2: M₂ = 7.2, SD₂ = 1.7, n₂ = 18

Compute 95% CI for the mean difference between the “learn how to learn” and “teach facts” groups.