Question

In: Statistics and Probability

You want to estimate the difference between the average grades on a certain math exam before...

You want to estimate the difference between the average grades on a certain math exam before the students take the associated math class and after they take the associated math class. You take the random samples of 5 students who have taken the class and 5 students who have not taken the class. You get the following results:

Not Taken Taken
54 82
25 76
73 98
23 43
42 38

A) Determine the population(s) and parameter(s) being discussed.

B) Determine which tool will help us find what we need (one sample z test, one sample t test, two sample t test, one sample z interval, one sample t interval, two sample t interval).

C) Check if the conditions for this tool hold.

D) Whether or not the conditions hold, use the tool you choose in part B. Use C=95% for all confidence intervals and alpha=5% for all significance tests.

* Be sure that all methods end with a sentence describing the results *

Solutions

Expert Solution

Given that,
null, H0: Ud = 0
alternate, H1: Ud != 0
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.776
since our test is two-tailed
reject Ho, if to < -2.776 OR if to > 2.776
we use Test Statistic
to= d/ (S/√n)
where
value of S^2 = [ ∑ di^2 – ( ∑ di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = 24
We have d = 24
pooled variance = calculate value of Sd= √S^2 = sqrt [ 4426-(120^2/5 ] / 4 = 19.66
to = d/ (S/√n) = 2.73
critical Value
the value of |t α| with n-1 = 4 d.f is 2.776
we got |t o| = 2.73 & |t α| =2.776
make Decision
hence Value of |to | < | t α | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 2.7297 ) = 0.0525
hence value of p0.05 < 0.0525,here we do not reject Ho
ANSWERS
---------------
A.
the difference between the average grades on a certain math exam before the students take the associated math class
and after they take the associated math class.
B.
Paired sample t test
C.
null, H0: Ud = 0
alternate, H1: Ud != 0
test statistic: 2.73
critical value: reject Ho, if to < -2.776 OR if to > 2.776
D.
decision: Do not Reject Ho
p-value: 0.0525
we do not have enough evidence to support the claim that take the random samples of 5 students who have taken the class and 5 students who have not taken the class


Related Solutions

The owner of a local supermarket wants to estimate the difference between the average number of...
The owner of a local supermarket wants to estimate the difference between the average number of gallons of milk sold per day on weekdays and weekends. The owner samples 8 weekdays and finds an average of 218.91 gallons of milk sold on those days with a standard deviation of 33.376. 10 (total) Saturdays and Sundays are sampled and the average number of gallons sold is 377.74 with a standard deviation of 49.365. Construct a 99% confidence interval to estimate the...
In order to estimate the difference between the average hourly wages of employees of two branches...
In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the following data have been gathered. Assume the degrees of freedom are 36. (Assume unequal variance) Sample 1 Sample 2 Downtown Store North Mall Store n 25 20 xbar $9 $9 s $2 $1 Q. What is a point estimate for the difference between the population means? Q. What is the critical value for a 95% confidence interval for the...
A travel website would like to estimate the difference between the average rental price of a...
A travel website would like to estimate the difference between the average rental price of a car with automatic transmission versus the average rental price of a car with manual transmission at a certain airport. The table below shows the average​ one-week rental prices for two random​ samples, as well as the population standard deviations and sample sizes for each type of car. Complete parts a and b. Sample mean Sample size Population standard deviation Automatic ​$411.90 54 ​$23 Manual...
Students who get more sleep get better grades. You want to estimate the hours of sleep...
Students who get more sleep get better grades. You want to estimate the hours of sleep per night that a college student gets. You select 29 students; on average they slept 6.2 hours with a standard deviation of 1.8 hours. The standard error is 0.334. In order to be 90% confident of an estimate of the population mean, what is the margin of error? (Enter the value with three decimal places, using form #.###)
Suppose you want to know if there is a difference in the average high temperatures in...
Suppose you want to know if there is a difference in the average high temperatures in February 2019 for three different areas. you record the following temperatures randomly for the three cities. A B C 18 -11 14 23 11 23 39 12 40 8 4 21 -4 -2 12 You decide to run a one-way ANOVA test on this data. A) write down the null and research hypotheses you are testing B) What are the F- statistic and P...
the average scores of math students in a certain school is 75 with a standard deviation...
the average scores of math students in a certain school is 75 with a standard deviation of 8.1. one hundred students were randomly selected, and the average score was found to be 71. the director wants to know wether students have deteriorated. significance level is 0.01 1) the hypotheses are ? 2)Decision and conclusion ? 3) the critical value is ? 4) the test statistic is ?
The score of a student on a certain exam is represented by a number between 0...
The score of a student on a certain exam is represented by a number between 0 and 1. Suppose that the student passes the exam if this number is at least 0.55. Suppose we model this experiment by a continuous random variable X, the score, whose probability density function is given by f(x) = { x if 0 <= x < 0.5 5x - 2 if 0.5 <= x <= 1 0 otherwise } a. What is the probability that...
1. Sample data on exam grades (Y), hours studied (X1) and homework average (X2) was used...
1. Sample data on exam grades (Y), hours studied (X1) and homework average (X2) was used to estimate the following regression equation: Y-hat = 60 + 5X1 + .1 X2 a. Interpret the value of the estimated constant (a) b. Interpret the estimated coefficient on X1 c. Interpret the estimated coefficient on X2 d. Predict the grade of a student who studied 5 hours and had a homework average of 90. Please answer questions and show work. Thanks
Suppose you want to estimate the average mpg for cars being driven from campus to the...
Suppose you want to estimate the average mpg for cars being driven from campus to the Brass Rail. You sample a few of cars. The average mpg in your sample is 22, and the standard deviation is 10. If you want the error of the confidence interval to be no greater than 0.3 and to be 90% likely to contain the true population mean, how many cars should you randomly sample? Group of answer choices A)about 3,007 cars B)about 9,112...
Suppose you monitor quality assurance for a local hospital and want to estimate the average length...
Suppose you monitor quality assurance for a local hospital and want to estimate the average length of stay (LOS) at your hospital. You take a random sample of 30 patients and find that the average LOS is 3.4 days with a sample standard deviation of 1.2 days. What is the 99% confidence interval for the population average length of stay?
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT