Question

Democrats and Republicans were surveyed on their support of gun control legislation using a standardized measure...

Democrats and Republicans were surveyed on their support of gun control legislation using a standardized measure (high scores representing greater support of gun control). Using the data provided, test the difference between the means (assume a two-tailed t-test) and answer the questions below:

Democrats                                                       Republicans

N = 50                                                              N = 40

Mean = 6.0                                                                  Mean = 4.9

Standard Deviation = 1.2                                 Standard Deviation = 1.4

8. In words: What would be the null hypothesis?

           

9. In words: What is the research hypothesis?

10. Calculate t: t = __________

11. What are your degrees of freedom? __________

12. Are your results significant at the .05 level? __________

13. In words: What are your conclusions about the null hypothesis?

Homework Answers

Answer #1

the null hypothesis: H0:

the research hypothesis Ha:
t =

t= (6-4.9)/sqrt((1.2)^2/50+(1.4)^2/40)

t= 1.1/sqrt(0.0288+0.049)

t= 1.1/sqrt(0.0778)

t= 1.1/0.279

t= 3.94

degrees of freedom is 77
The p-value is .000178.The result is significant because p < .05.

onclusions about the null hypothesis: REJECT NULL HYPOTHESIS H0.

Conclusion: We have sufficient evidence to show that there is difference between two means.

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Democrats and Republicans were surveyed on their support of gun control legislation using a standardized measure...
Democrats and Republicans were surveyed on their support of gun control legislation using a standardized measure (high scores representing greater support of gun control). Using the data provided, calculate t. Democrats Republicans N= 50 N= 50 Mean = 6.0 Mean = 4.9 Std. Dev. = 1.2 Std. Dev. 1.4
To test the effectiveness of a studying program, participants are randomly assigned to either the control...
To test the effectiveness of a studying program, participants are randomly assigned to either the control group or the treatment group (N=10 for each). At the end of the trials, the average test score increase in the control group is 3.5 points (s=0.7) and the treatment group increased 1.9 (s=1.1). Test the null hypothesis that there is no difference in test score improvement between the treatment and control groups (alpha = 0.05). For independent sample t-tests: mean values for each...
A researcher knows that scores on a standardized language measure are normally distributed with a μ...
A researcher knows that scores on a standardized language measure are normally distributed with a μ = 50 and σ = 20. She wants to know if an online computer program will improve scores on the standardized test. She gives a sample of n = 25 students the study program and calculates the sample mean (M = 60) for her group of students that received the study program. She calculates her test statistic to be z = 2.5 ! She...
Using the following table to determine if the following are significant: a) t=2.1211 N=9 b) t=3.813...
Using the following table to determine if the following are significant: a) t=2.1211 N=9 b) t=3.813 N=6 c. State the null hypothesis when comparing a control and treated group. d. What is a standard deviation? e. How is the standard deviation depicted on a graph?
4) 25 randomly selected people were surveyed and it was found that the average amount they...
4) 25 randomly selected people were surveyed and it was found that the average amount they charged was 1200 in a year with a standard deviation of 600. Given α = 0.025, test the claim that the mean amount charged by all people with credit cards is greater than 1300. Confirm your results using the p–value method as well as the critical t value. 5) 5) In a study 800 people out of 1000 people surveyed indicated that they supported...
In a consumer research study, several Meijer and Walmart stores were surveyed at random and the...
In a consumer research study, several Meijer and Walmart stores were surveyed at random and the average basket price was recorded for each. You wish to determine if the average basket price for Meijer is different from the average basket price for Walmart. It was found that the average basket price for 18 randomly chosen Meijer stores (group 1) was $49.451 with a standard deviation of $12.3146. Similarly, a random sample of 25 Walmart stores (group 2) had an average...
(Using Excel) A producer of computer aided design software has received numerous calls for technical support....
(Using Excel) A producer of computer aided design software has received numerous calls for technical support. The company has a service standard of 4 days for the mean resolution time for questions. The manager of the technical support group has been receiving some complaints about the resolution being too long.   During one week a sample of 40 customer calls result in a sample mean of 5.25 days and a standard deviation of 13.5 days. 1. Set up the hypothesis test?   ...
California had stricter gun laws than Texas. However, California had a greater proportion of gun murders...
California had stricter gun laws than Texas. However, California had a greater proportion of gun murders than Texas. Here we test whether or not the proportion was significantly greater in California. A significant difference is one that is unlikely to be a result of random variation. The table summarizes the data for each state. The p̂'s are actually population proportions but you should treat them as sample proportions. The standard error (SE) is given to save calculation time if you...
Using the table below, would you please check my answers. The answer for the p-value is...
Using the table below, would you please check my answers. The answer for the p-value is supposed to be 0.0000; however, I keep getting 0.9999999977 as the answer. How do you find the correct p-value? Where am I going wrong in my calculation? Regardless of your answer to part (a) using R, run a t-test for the difference between the two treatment means at the α = 5% level of significance. Specifically, we are interested in knowing if there exists...
The U.S. Center for Disease Control reports that the mean life expectancy was 47.6 years for...
The U.S. Center for Disease Control reports that the mean life expectancy was 47.6 years for whites born in 1900 and 33.0 years for nonwhites. Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124 whites, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 nonwhites, the mean life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis...