CNNBC recently reported that the mean annual cost of auto insurance is 988 dollars. Assume the standard deviation is 105 dollars. You will use a simple random sample of 110 auto insurance policies.

Find the probability that a single randomly selected policy has a mean value between 994 and 1008 dollars.
P(994 < X < 1008) =

Find the probability that a random sample of size n=110n=110 has a mean value between 994 and 1008 dollars.
P(994 < M < 1008) =

Many different manufacturers sell residential gas ranges. The cost in dollars of four gas ranges is given below.

529 664 709 800

1. suppose a random sample of size two is selected from this population without replacement. Find the sampling distribution of the sample mean.
2. Suppose a random sample of size is selected from this population with replacement. Find the sampling distribution of the sample mean
3. How are these two distributions similar? How are they different?

Based on historical data, your manager believes that 44% of the company's orders come from first-time customers. A random sample of 141 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is between 0.26 and 0.48?

Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations. Answer = (Enter your answer as a number accurate to 4 decimal places.)

In a study of 1910 schoolchildren in Australia, 1050 children indicated that they normally watch TV before school in the morning. (Interestingly, only 35% of the parents said their children watched TV before school!)

(a)

Construct a 95% confidence interval for the true proportion of Australian children who say they watch TV before school. (Round your answers to three decimal places.)

(_____,________)
What assumption about the sample must be true for the method used to construct the interval to be valid?(b)

The 1910 schoolchildren used in the study formed a random sample from the population of children in Australia who normally watch TV before school in the morning.

The 1050 children who indicated that they normally watch TV before school in the morning formed a random sample from the population of schoolchildren in Australia.    

The 1910 schoolchildren used in the study formed a random sample from the population of schoolchildren in Australia.

The 1050 children who indicated that they normally watch TV before school in the morning formed a random sample from the population of children in Australia who normally watch TV before school in the morning.

How much money do people spend on graduation gifts? In 2007, a federation surveyed 2415 consumers who reported that they bought one or more graduation gifts that year. The sample was selected in a way designed to produce a sample representative of adult Americans who purchased graduation gifts in 2007. For this sample, the mean amount spent per gift was $58.15. Suppose that the sample standard deviation was $20. Construct a 98% confidence interval for the mean amount of money spent per graduation gift in 2007. (Round your answers to three decimal places.)
(  ,  )

Interpret the interval.

We are 98% confident that the mean amount of graduation money spent was within this interval.

We are 98% confident that the mean amount of money spent per graduation gift in 2007 was within this interval.    

We are confident that the mean amount of money spent per graduation gift in 2007 was within this interval 98% of the time.

We are confident that 98% of the amount of money spent per graduation gift in 2007 was within this interval.

Assume that a researcher is interested in the relationship between hours of sleep and anxiety. Eight individuals are randomly selected and the amount of hours slept and anxiety scores are measured. The scores are reported in the following table. Calculate the correlation coefficient. Use α = .05 to conduct a hypothesis test on correlation.

Hours of sleep	Anxiety score
5.5	55
6	47
8	45
7.25	50
8.5	35
7	39
8.75	39
9	36

*SHOW WORK FOR THIS PROBLEM

1) H₁: ρ  , H₀: ρ  

2) r critical value =  *DO NOT ROUND

3) r =  *ROUND TO FOUR DECIMALS

4)  (RTN or FTR)

5) What is the effect size for this relationship?  *ROUND TO FOUR DECIMAL

For the following, Use the five-step approach to hypothesis testing found on page 8-16. It states. You can use excel to compute the data or you can do it by hand. The youtube videos provided in the links will walk you through the steps to complete the following problems.

1. State the hypothesis and identify the claim.

H0:

H1:

2. Use a Pearson Correlation
3. Compute the test Value It will be a correlation
4. Make a decision to reject or fail to reject the null hypothesis.
5. Summarize the results

Problem #3 You are a researcher who wants to know if there is a relationship between variable Y and variable X. You hypothesize that there will be a strong positive relationship between variable Y GPA and Variable X hours of sleep. After one semester, you select five students at random out of 200 students who have taken a survey and found that they do not get more than 5 hours of sleep per night. You select five more students at random from the same survey that indicates students getting at least seven hours of sleep per night. You want to see if there is a relationship between GPA and hours of sleep. Using a Pearson Product Correlation Coefficient statistic, determine the strength and direction of the relationship and determine if you can reject or fail to reject the HO:

Variable Y        Variable X    

2.5                      5               

3.4                      8                

2.0                      4               

2.3                     4.5              

1.6                     3     

3.2                     6

2.8                     7

3.5                     7.5

4.0                     6.5

3.8                     7

solve it with exact data given not an example or other illustration.

A researcher believes that alcohol intoxication might severely impair driving ability. To test this, she subjects 10 volunteers to a driving simulation test, first when sober, and then, after drinking amounts sufficient to raise their blood alcohol to .04. The researcher measures performance as the number of simulated obstacles with which the driver collides. Thus, the higher the number, the poorer the driving. The data is in the Excel file in the tab labeled Question 4. Test whether there are differences before and after drinking. Conduct a t-test: Two-Sample for Means.

Before Drinking:

1
2
0
0
2
1
4
0
1
2
0

After drinking:

4
2
1
2
5
3
3
2
4
3
1

a. What is the null hypothesis?

b. What is the research hypothesis?

c. Why run a Two-Sample for Means t-test?

d. Interpret the findings. What are the results of the hypothesis test? Can you reject the null hypothesis?

Below are results from a regression analysis. The dependent variable is the percent vote for Woodrow Wilson in 1916, measured from 0 to 100. Each observation is a New Jersey county, of which there are 20. The independent variables are the county’s percentage of Democrats (0 to 100), percentage working class (0 to 100), and the number of shark attacks that occurred in it in 1916 (0, 1, 2, . . . ). Shark attacks are rare: no county experienced more than 3.

	estimate
(Intercept)	0.327
% Democrats	0.897
% workers	-0.121
# shark attacks	-0.506

(a) How do we interpret the coefficient estimate for # shark attacks? (b) What would be the predicted vote share for Wilson in a county that is 40% Democratic, has 80% working class, and experienced no shark attacks? What about if the same county experienced 1 shark attack?

Please note that for all problems in this course, the standard cut-off (alpha) for a test of significance will be .05, and you always report the exact power unless SPSS output states p=.000 (you’d report p<.001). Also, remember when hand-calculating, always use TWO decimal places so that deductions in grading won’t be due to rounding differences.

Problem Set 1: A child trauma counselor knows that for her region, 15% of children are in extreme poverty, 50% are below poverty, and 35% of the children are above the poverty level. She wants to know if her services reaches a similar demographic to what is known in the region. Services provided at each demographic level for her 36 clients are provided in the table below. Hint: Review Week 4.    Extreme poverty Below poverty Above poverty 8 15 13

1. Paste appropriate SPSS output. (2 pts)

2. Paste appropriate SPSS graph. (2 pts)

3. Write an APA-style Results section based on your analysis. All homework “Results sections” should follow the examples provided in the presentations and textbooks. They should include the statistical statement within a complete sentence that mentions the type of test conducted, whether the test was significant, and if relevant, effect size and/or post hoc analyses. Don’t forget to include a decision about the null hypothesis. (2 pts)

Researchers would like to see if there is a difference in satisfaction levels in a group of families’ use of service centers following a social service intervention. The data for this study is in the Excel data file, under the tab labeled Question 3. Conduct the t-test: Paired with Two-Sample for Means.

Before intervention:

1.3
2.5
2.3
8.1
5
7
7.5
5.2
4.4
7.6
9
7.6
4.5
1.1
5.6
6.2
7
6.9
5.6
5.2

After intervention:

6.5
8.7
9.8
10.2
7.9
6.5
8.7
7.9
8.7
9.1
8.4
6.4
7.2
5.8
6.9
5.9
7.6
7.8
7.3
4.6

a. What is the null hypothesis?

b. What is the research hypothesis?

c. Why run a Two-Sample for Means t-test?

d. Interpret the findings. What are the results of the hypothesis test? Can you reject the null hypothesis?

1. A company produces electric motors for use in home appliances. One of the company’s production managers is interested in examining the relationship between the dollars spent per month in inspecting finished motor products (X) and the number of motors produced during that month that were returned by dissatisfied customers (Y). He has collected the data in the file P11_32.xlsx to explore this relationship for the past 36 months.

a. Estimate a simple linear regression equation using the given data.

If required, round your answers to three decimal places. For subtractive or negative numbers use a minus sign even if there is a + sign before the blank. (Example: -300)

Q1: The predicted change in number of motors returned that is associated with an increase of $1,000 in inspection expenditures: (_____________)

Month	Inspection Expenditures	Motors Returned
1	$43,549	67
2	$55,831	64
3	$48,529	65
4	$48,551	65
5	$69,286	65
6	$64,390	64
7	$58,323	64
8	$76,902	67
9	$65,976	64
10	$43,437	67
11	$41,495	67
12	$55,067	64
13	$60,743	64
14	$41,061	68
15	$64,578	64
16	$70,219	65
17	$68,254	65
18	$50,930	65
19	$76,354	67
20	$43,927	67
21	$60,250	64
22	$42,219	67
23	$79,183	68
24	$66,628	64
25	$79,675	68
26	$52,793	65
27	$50,579	65
28	$66,856	64
29	$42,954	67
30	$62,449	64
31	$42,732	67
32	$78,455	67
33	$74,487	66
34	$62,685	64
35	$74,411	66
36	$42,407	67

A corrections researcher is interested in gender differences in the amount of bail that arrestees have to pay. The data for this study is in the Excel data file, under the tab labeled Question 2. Conduct a t-test: Two-Sample Assuming Equal Variances.

Men: 1000,300,1500,200,500,150,1000,500,1200

Females: 500, 250, 100,200, 200, 250,100,150,100

a. What is the null hypothesis?

b. What is the research hypothesis?

c. Why run a Two-Sample Assuming Equal Variances t-test?

d. Interpret the findings. What are the results of the hypothesis test? Can you reject the null hypothesis?

Assignment #3: Inferential Statistics Analysis and Writeup

Part A: Inferential Statistics Data Analysis Plan and Computation

Introduction: I chose to imagine I am a 36 year old married individual with a large family. (UniqueID#30)

Variables Selected:

Table 1: Variables Selected for Analysis

Variable Name in the Data Set	Variable Type	Description	Qualitative or Quantitative
Variable 1: Marital Status	Socioeconomic	Marital Status of Head of Household	Qualitative
Variable 2: Housing	Expenditure	Total Amount of Annual Expenditure on Housing	Quantitative
Variable 3: Transport	Expenditure	Total Amount of Annual Expenditure on Transportation	Quantitative

Data Analysis:

1. Confidence Interval Analysis: For one expenditure variable, select and run the appropriate method for estimating a parameter, based on a statistic (i.e., confidence interval method) and complete the following table (Note: Format follows Kozak outline):

Table 2: Confidence Interval Information and Results

Name of Variable:

State the Random Variable and Parameter in Words:

Confidence interval method including confidence level and rationale for using it:

State and check the assumptions for confidence interval:

Method Used to Analyze Data:

Find the sample statistic and the confidence interval:

Statistical Interpretation:

2. Hypothesis Testing: Using the second expenditure variable (with socioeconomic variable as the grouping variable for making two groups), select and run the appropriate method for making decisions about two parameters relative to observed statistics (i.e., two sample hypothesis testing method) and complete the following table (Note: Format follows Kozak outline):

Table 3: Two Sample Hypothesis Test Analysis

Research Question:

Two Sample Hypothesis Test that Will Be Used and Rationale for Using It:

State the Random Variable and Parameters in Words:

State Null and Alternative Hypotheses and Level of Significance:

Method Used to Analyze Data:

Find the sample statistic, test statistic, and p-value:

Conclusion Regarding Whether or Not to Reject the Null Hypothesis:

Part B: Results Write Up

Confidence Interval Analysis:

Two Sample Hypothesis Test Analysis:

Discussion:

Data Set:

UniqueID#	SE-MaritalStatus	SE-Income	SE-AgeHeadHousehold	SE-FamilySize	USD-AnnualExpenditures	USD-Food	USD-Housing	USD-Transport
1	Not Married	95432	51	1	55120	7089	18391	115
2	Not Married	97469	35	4	54929	6900	18514	145
3	Not Married	96664	53	3	55558	7051	18502	168
4	Not Married	96653	51	4	56488	6943	18838	124
5	Not Married	94867	60	1	55512	6935	18633	131
6	Not Married	97912	49	1	55704	6937	18619	152
7	Not Married	96886	44	2	55321	6982	18312	153
8	Not Married	96244	56	4	56051	7073	18484	141
9	Not Married	95366	48	2	57082	7130	18576	149
10	Not Married	96727	39	2	56440	7051	18376	120
11	Not Married	96697	49	2	56453	6971	18520	136
12	Not Married	95744	52	4	55963	7040	18435	146
13	Not Married	96572	59	2	56515	7179	18648	123
14	Not Married	98717	40	3	56393	7036	18389	114
15	Not Married	94929	59	2	55247	6948	18483	133
16	Married	95778	42	4	73323	9067	22880	201
17	Married	109377	48	4	83530	10575	23407	99
18	Married	95706	52	4	71597	8925	22376	181
19	Married	95865	46	1	74789	9321	22621	168
20	Married	109211	42	4	82503	11566	22219	62
21	Married	95994	55	4	73404	9231	22852	177
22	Married	114932	44	5	81186	11077	26411	153
23	Married	112559	39	3	80934	11189	25531	73
24	Married	95807	56	4	72949	9210	23139	186
25	Married	99610	36	2	73550	9513	27164	33
26	Married	95835	54	3	73092	9111	23252	186
27	Married	102081	42	4	82331	11738	23374	121
28	Married	104671	41	4	82786	10420	22245	84
29	Married	107028	46	4	82816	10840	25671	109
30	Married	114505	36	5	78325	11375	26006	140

In a survey conducted to determine, among other things, the cost of vacations, 64 individuals were randomly sampled. Each person was asked to compute the cost of her or his most recent vacation. The sample showed a sample mean of $1810. Assuming that the population standard deviation σ is $600, construct a 90% confidence interval for the average cost of all vacations. (Round your final answer to 3 decimal places).