Questions
Answer each of the following questions by typing the answers into the Assignment text box below....

Answer each of the following questions by typing the answers into the Assignment text box below. For the predicted value of Y we will use Y-hat.

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.

The mean grade was 78.5. The standard error of estimate is Se = 5.25 and the coefficient of determination is R2=.873.

a. Interpret the standard error of the estimate.

b. Interpret the coefficient of determination.

c. Is this estimated regression equation a good fit? EXPLAIN.

Please show work.

Thanks

In: Statistics and Probability

a)When is it appropriate to calculate a z-test? b)What is the formula for a z-test? c)If...

a)When is it appropriate to calculate a z-test?

b)What is the formula for a z-test?

c)If we set α = 0.10 what does that tell us?

d)If p < α then what should you do?

e)What is the z-table used for when conducting a z-test?

f)What are the 7 steps for hypothesis testing?

In: Statistics and Probability

write a null alternative hypotheses that demonstrate an understanding of the guiding principles of ANOVA. explain...

write a null alternative hypotheses that demonstrate an understanding of the guiding principles of ANOVA. explain how you came up with your answer.

In: Statistics and Probability

A soda bottling plant fills cans labeled to contain 12 ounces of soda. The filling machine...

A soda bottling plant fills cans labeled to contain 12 ounces of soda. The filling machine varies and does not fill each can with exactly 12 ounces. To determine if the filling machine needs adjustment, each day the quality control manager measures the amount of soda per can for a random sample of 50 cans. Experience shows that its filling machines have a known population standard deviation of 0.35 ounces.

In today's sample of 50 cans of soda, the sample average amount of soda per can is 12.1 ounces. Construct and interpret a 90% confidence interval estimate for the true population average amount of

soda contained in all cans filled today at this bottling plant. Use a 90% confidence level.
X =

population parameter:  =

Random Variable X=

In: Statistics and Probability

Question 1 The Highway Safety Department wants to study the driving habits of individuals. A sample...

Question 1
The Highway Safety Department wants to study the driving habits of individuals. A sample of 41
cars traveling on the highway revealed an average speed of 60 miles per hour and a standard
deviation of 7 miles per hour. The population of car speeds is approximately normally distributed.
Determine a 90% confidence interval estimate for the speed of all cars.

In: Statistics and Probability

Let x = age in years of a rural Quebec woman at the time of her...

Let x = age in years of a rural Quebec woman at the time of her first marriage. In the year 1941, the population variance of x was approximately σ2 = 5.1. Suppose a recent study of age at first marriage for a random sample of 41 women in rural Quebec gave a sample variance s2 = 2.9. Use a 5% level of significance to test the claim that the current variance is less than 5.1. Find a 90% confidence interval for the population variance.

(a) What is the level of significance?


State the null and alternate hypotheses.

Ho: σ2 = 5.1; H1: σ2 > 5.1
Ho: σ2 < 5.1; H1: σ2 = 5.1    
Ho: σ2 = 5.1; H1: σ2 ≠ 5.1
Ho: σ2 = 5.1; H1: σ2 < 5.1


(b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)


What are the degrees of freedom?


What assumptions are you making about the original distribution?

We assume a exponential population distribution.
We assume a normal population distribution.    
We assume a uniform population distribution.
We assume a binomial population distribution.


(c) Find or estimate the P-value of the sample test statistic.

P-value > 0.100
0.050 < P-value < 0.100    
0.025 < P-value < 0.050
0.010 < P-value < 0.025
0.005 < P-value < 0.010
P-value < 0.005


(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?

Since the P-value > α, we fail to reject the null hypothesis.
Since the P-value > α, we reject the null hypothesis.    
Since the P-value ≤ α, we reject the null hypothesis.
Since the P-value ≤ α, we fail to reject the null hypothesis.


(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, there is insufficient evidence to conclude that the variance of age at first marriage is less than 5.1.
At the 5% level of significance, there is sufficient evidence to conclude that the that the variance of age at first marriage is less than 5.1.    


(f) Find the requested confidence interval for the population variance. (Round your answers to two decimal places.)

lower limit
upper limit    


Interpret the results in the context of the application.

We are 90% confident that σ2 lies outside this interval.
We are 90% confident that σ2 lies below this interval.    
We are 90% confident that σ2 lies within this interval.
We are 90% confident that σ2 lies above this interval.

In: Statistics and Probability

7. It is widely accepted that people are a little taller in the morning than at...

7. It is widely accepted that people are a little taller in the morning than at night. Here we perform a test on how big the difference is. In a sample of 30adults, the morning height and evening height are given in millimeters (mm) in the table below. Use this data to test the claim that, on average, people are more than 10 mm taller in the morning than at night. Test this claim at the 0.01 significance level.



(a) The claim is that the mean difference (x - y) is more than 10 mm (μd > 10). What type of test is this?

This is a left-tailed test.This is a right-tailed test.     This is a two-tailed test.


(b) What is the test statistic? Round your answer to 2 decimal places.
t

d

=  

(c) Use software to get the P-value of the test statistic. Round to 4 decimal places.
P-value =  

(d) What is the conclusion regarding the null hypothesis?

reject H0fail to reject H0     


(e) Choose the appropriate concluding statement.

The data supports the claim that, on average, people are more than 10 mm taller in the morning than at night.There is not enough data to support the claim that, on average, people are more than 10 mm taller in the morning than at night.     We reject the claim that, on average, people are more than 10 mm taller in the morning than at night.We have proven that, on average, people are more than 10 mm taller in the morning than at night.

    
    
AM Height (x) PM Height (y) (x - y)
1609 1600 9
1770 1757 13
1563 1556 7
1599 1588 11
1659 1652 7
1561 1548 13
1621 1607 14
1401 1390 11
1440 1430 10
1693 1680 13
1467 1453 14
1667 1655 12
1736 1723 13
1424 1409 15
1785 1774 11
1585 1567 18
1472 1459 13
1629 1609 20
1409 1398 11
1481 1471 10
1589 1580 9
1499 1487 12
1532 1524 8
1507 1498 9
1445 1439 6
1637 1629 8
1658 1644 14
1680 1674 6
1799 1790 9
1436 1423 13

In: Statistics and Probability

The age distribution of the Canadian population and the age distribution of a random sample of...

The age distribution of the Canadian population and the age distribution of a random sample of 455 residents in the Indian community of a village are shown below.

Age (years) Percent of Canadian Population Observed Number in the Village

Under 5 7.2% 42

5 to 14 13.6% 77

15 to 64 67.1% 288

65 and older 12.1% 48

Use a 5% level of significance to test the claim that the age distribution of the general Canadian population fits the age distribution of the residents of Red Lake Village.

(a) What is the level of significance?

State the null and alternate hypotheses.

H0: The distributions are different. H1: The distributions are different.

H0: The distributions are the same. H1: The distributions are the same.

H0: The distributions are the same. H1: The distributions are different.

H0: The distributions are different. H1: The distributions are the same.

(b) Find the value of the chi-square statistic for the sample. (Round your answer to three decimal places.)

Are all the expected frequencies greater than 5? Yes No

What sampling distribution will you use?

normal uniform chi-square binomial Student's t

What are the degrees of freedom?

(c) Estimate the P-value of the sample test statistic.

P-value > 0.100

0.050 < P-value < 0.100

0.025 < P-value < 0.050

0.010 < P-value < 0.025

0.005 < P-value < 0.010

P-value < 0.005

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis that the population fits the specified distribution of categories?

Since the P-value > α, we fail to reject the null hypothesis.

Since the P-value > α, we reject the null hypothesis.

Since the P-value ≤ α, we reject the null hypothesis.

Since the P-value ≤ α, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, the evidence is insufficient to conclude that the village population does not fit the general Canadian population.

At the 5% level of significance, the evidence is sufficient to conclude that the village population does not fit the general Canadian population.

In: Statistics and Probability

The following table shows ceremonial ranking and type of pottery sherd for a random sample of...

The following table shows ceremonial ranking and type of pottery sherd for a random sample of 434 sherds at an archaeological location.

Ceremonial Ranking Cooking Jar Sherds Decorated Jar Sherds (Noncooking) Row Total
A 86 49 135
B 95 50 145
C 77 77 154
Column Total 258 176 434

Use a chi-square test to determine if ceremonial ranking and pottery type are independent at the 0.05 level of significance.

(a) What is the level of significance?


State the null and alternate hypotheses.

H0: Ceremonial ranking and pottery type are not independent.
H1: Ceremonial ranking and pottery type are independent.

H0: Ceremonial ranking and pottery type are not independent.
H1: Ceremonial ranking and pottery type are not independent.    

H0: Ceremonial ranking and pottery type are independent.
H1: Ceremonial ranking and pottery type are independent.

H0: Ceremonial ranking and pottery type are independent.
H1: Ceremonial ranking and pottery type are not independent.


(b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to at least three decimal places. Round the test statistic to three decimal places.)


Are all the expected frequencies greater than 5?

Yes No    


What sampling distribution will you use?

uniform
Student's t    
chi-square
normal
binomial


What are the degrees of freedom?


(c) Find or estimate the P-value of the sample test statistic. (Round your answer to three decimal places.)

p-value > 0.100
0.050 < p-value < 0.100    
0.025 < p-value < 0.050
0.010 < p-value < 0.025
0.005 < p-value < 0.010
p-value < 0.005


(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence?

Since the P-value > α, we fail to reject the null hypothesis.
Since the P-value > α, we reject the null hypothesis.    
Since the P-value ≤ α, we reject the null hypothesis.
Since the P-value ≤ α, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, there is sufficient evidence to conclude that ceremonial ranking and pottery type are not independent.
At the 5% level of significance, there is insufficient evidence to conclude that ceremonial ranking and pottery type are not independent.    

In: Statistics and Probability

Suppose that an accounting firm does a study to determine the time needed to complete one...

Suppose that an accounting firm does a study to determine the time needed to complete one persons tax form. It randomly surveys 100 people. The sample mean is 23.6 hours. There is a known standard deviation of 7.0 hours. The population distribution is assumed to be normal. Construct a 90% confidence interval for the population mean time to complete the tax forms.

In: Statistics and Probability

Grades and AM/PM Section of Stats: There were two large sections of statistics this term at...

Grades and AM/PM Section of Stats: There were two large sections of statistics this term at State College, an 8:00 (AM) section and a 1:30 (PM) section. The final grades for both sections are summarized in the contingency table below.

Observed Frequencies: Oi's

A B C D F Totals
AM 6 11 17 18 20 72
PM 19 19 17 12 9 76
Totals 25 30 34 30 29 148

The Test: Test for a significant dependent relationship between grades and the section of the course. Conduct this test at the 0.05 significance level.(a) What is the null hypothesis for this test?

H0: The section (AM/PM) of a course and the grades are independent variables.

H0: The section (AM/PM) of a course and the grades are dependent variables.     


(b) What is the value of the test statistic? Round to 3 decimal places unless your software automatically rounds to 2 decimal places.

χ2 = ?

(c) Use software to get the P-value of the test statistic. Round to 4 decimal places unless your software automatically rounds to 3 decimal places.
P-value = ?

(d) What is the conclusion regarding the null hypothesis?

reject H0

fail to reject H0    


(e) Choose the appropriate concluding statement.

We have proven that grades and section of the course are independent.

The evidence suggests that there is a significant dependent relationship between grades and the section of the course.    

There is not enough evidence to conclude that there is a significant dependent relationship between grades and the section of the course

In: Statistics and Probability

What will the probability always be of events occurring that are mutually exclusive?  What are the...

What will the probability always be of events occurring that are mutually exclusive?

 What are the different types of probability distribution and what are their characteristics?

In: Statistics and Probability

probability 10%. n= 50 find between 15 and 18 times.

probability 10%. n= 50 find between 15 and 18 times.

In: Statistics and Probability

Group A Group B Subject 1 40 30 Subject 2 42 45 Subject 3 30 38...

Group A Group B
Subject 1 40 30
Subject 2 42 45
Subject 3 30 38
Subject 4 37 32
Subject 5 23 28

A small hypothetical study explored the response to two treatments where each subject in the study received both treatments (e.g. in a cross-over design). The data are given below 1) Analyze the information using a paired t-test (e.g. with GraphPad) a) What is the observed value of the test statistic from this test? b) What can you conclude about difference in mean response to the two treatments? Why?

In: Statistics and Probability

A STAT 200 instructor believes that the average quiz score is a good predictor of final...

A STAT 200 instructor believes that the average quiz score is a good predictor of final exam score. A random sample of 10 students produced the following data where x is the average quiz score and y is the final exam score.

x- 80, 95, 50, 60, 100, 55, 85, 70, 75, 85

y - 70, 96, 50, 63, 96, 60, 83, 60, 77, 87

(a) Find an equation of the least squares regression line. Round the slope and y-intercept value to two decimal places. Describe method for obtaining results. Show all work; writing the correct equation, without supporting work, will receive no credit. (b) Based on the equation from part (a), what is the predicted final exam score if the average quiz score is 65? Show all work and justify your answer. (c) Based on the equation from part (a), what is the predicted final exam score if the average quiz score is 40? Show all work and justify your answer. (d) Which predicted final exam score that you calculated for (b) and (c) do you think is closer to the true final exam score and why?

In: Statistics and Probability