Question

Question 1

The American Pet Products Association conducted a survey in 2011 and determined that 60% of dog owners have only one dog, 28% have two dogs, and 12% have three or more. Supposing that you have decided to conduct your own survey and have collected the data below, determine whether your data supports the results of the APPA study. Use a significance level of 0.05.

Data: Out of 119 dog owners, 70 had one dog and 30 had two dogs.

What is the p-value when testing whether your data supports the results of the APPA study? (note: The question asked has been updated)

Question 1 options:

	0.2316
	0.2710
	0.3832
	0.4487

Question 2

A survey of 325 middle school students from a city school district asks, among other things, for students’ gender, whether they prefer cats or dogs, and whether they prefer basketball or football. With these data, we can investigate whether, in this city, girls prefer cats over dogs, whether gender matters in terms of favorite sport, and whether there is a relationship between favorite pet and favorite sport.

Here are part of the data:

Pet	Boy	Girl
Cat	51	69
Dog	110	95

Suppose that you plan to use a chi-square test for determining whether there is any gender gap in the preference of certain pets (cats/dogs).

What is the p-value?

Question 2 options:

	0.0314
	0.0405
	0.0522
	0.0689

Question 3

Suppose we have survey data for 1,000 randomly selected local pet owners. We wish to determine if the population proportion of local ferret owners is different from the national average of 6.5%. Out of the 1000 selected pet owners, 85 are ferret owners.

Test whether there is a significant difference between the local ferret rate and the national average.

The p-value is

Question 3 options:

	0.0045
	0.0078
	0.0103
	0.0351

Question 4

Suppose we have survey data for 1,000 randomly selected local pet owners. We wish to determine if the population proportion of local ferret owners is different from the national average of 6.5%. Out of the 1000 selected pet owners, 85 are ferret owners.

Give a 95% confidence interval for the proportion of local ferret owners.

Question 4 options:

	0.04 to 0.13
	0.07 to 0.10
	0.05 to 0.12
	0.06 to 0.11

Question 5

A high school math teacher believes that male and female students who graduated from the school perform equally well on SAT math test. She randomly chooses 10 male students and 10 female students who graduated from this school. The following are the SAT math scores of the 20 students:

Male: 23, 30, 27, 29, 22, 34, 36, 28, 28, 31
Female: 22, 33, 30, 28, 28, 31, 34, 25, 29, 21

Test the teacher's claim.

What is the p-value?

Question 5 options:

	0.6268
	0.0407
	0.7223
	0.0521

Answer 1

1)

observed frequencey, O	expected proportion	expected frequency,E			(O-E)²/E
70	0.600	71.40			0.027
30	0.280	33.32			0.331
19	0.120	14.28			1.560

chi square test statistic,X² = Σ(O-E)²/E =   1.918              
                  
level of significance, α=   0.05              
Degree of freedom=k-1=   3   -   1   =   2
                  
P value =   0.3832   [ excel function: =chisq.dist.rt(test-stat,df) ]          

option (c)

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2)

Chi-Square Test of independence
	Observed Frequencies
	0
0	boy	girl					Total
cat	51	69					120
dog	110	95					205
Total	161	164					325
	Expected frequency of a cell = sum of row*sum of column / total sum
	Expected Frequencies
	boy	girl					Total
cat	161*120/325=59.446	164*120/325=60.554					120
dog	161*205/325=101.554	164*205/325=103.446					205
Total	161	164					325
	(fo-fe)^2/fe
cat	1.200	1.178
dog	0.702	0.690

Chi-Square Test Statistic,χ² = Σ(fo-fe)^2/fe =   3.770
  
Level of Significance =   0.05
Number of Rows =   2
Number of Columns =   2
Degrees of Freedom=(#row - 1)(#column -1) = (2- 1 ) * ( 2- 1 ) =   1
  
p-Value =   0.0522

option (c)

.................

3)

Ho :   p =    0.065          
H1 :   p ╪   0.065       (Two tail test)  
                  
Number of Items of Interest,   x =   85          
Sample Size,   n =    1000          
                  
Sample Proportion ,    p̂ = x/n =    0.0850          
                  
Standard Error ,    SE = √( p(1-p)/n ) =    0.0078          
Z Test Statistic = ( p̂-p)/SE = (   0.0850   -   0.065   ) /   0.0078
                        
p-Value   =   0.0103   [excel formula =2*NORMSDIST(z)]      

option (c)

.................

4)

Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   85          
Sample Size,   n =    1000          
                  
Sample Proportion ,    p̂ = x/n =    0.085          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0088          
margin of error , E = Z*SE =    1.960   *   0.0088   =   0.0173
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.085   -   0.0173   =   0.0677
Interval Upper Limit = p̂ + E =   0.085   +   0.0173   =   0.1023
                  
95%   confidence interval is (   0.07 < p <    0.1 )

option (b)

.........

5)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
  
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   28.80                  
standard deviation of sample 1,   s1 =    4.34                  
size of sample 1,    n1=   10                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   28.10                  
standard deviation of sample 2,   s2 =    4.33                  
size of sample 2,    n2=   10                  
                          
difference in sample means =    x̅1-x̅2 =    28.8000   -   28.1   =   0.70  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    4.3365                  
std error , SE =    Sp*√(1/n1+1/n2) =    1.9394                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   0.7000   -   0   ) /    1.94   =   0.361
                          
Degree of freedom, DF=   n1+n2-2 =    18                  
  
p-value =        0.7223   (excel function: =T.DIST.2T(t stat,df) )              

option (c)

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thanks

done all question

please upvote

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