Question

CH10.

1. Consider the following hypothesis test.

The following results are for two independent samples taken from the two populations.

Sample 1	Sample 2
n 1 = 80	n 2 = 70
x 1 = 104	x 2 = 106
σ 1 = 8.4	σ 2 = 7.6

1. What is the value of the test statistic? If required enter negative values as negative numbers (to 2 decimals).
2. What is the p-value (to 4 decimals)? Use z-table.

2.

The USA Today reports that the average expenditure on Valentine's Day is $100.89. Do male and female consumers differ in the amounts they spend? The average expenditure in a sample survey of 46 male consumers was $135.67, and the average expenditure in a sample survey of 34 female consumers was $68.64. Based on past surveys, the standard deviation for male consumers is assumed to be $38, and the standard deviation for female consumers is assumed to be $20.

1. What is the point estimate of the difference between the population mean expenditure for males and the population mean expenditure for females (to 2 decimals)?
   
2. At 99% confidence, what is the margin of error (to 2 decimals)?
   
3. Develop a 99% confidence interval for the difference between the two population means (to 2 decimals)

3.

Forbes reports that women trust recommendations from Pinterest more than recommendations from any other social network platform (Forbes website, April 10, 2012). But, does trust in Pinterest differ by gender? The following sample data show the number of women and men who stated in a recent sample that they trust recommendations made on Pinterest.

	Women	Men
Sample	150	170
Trust Recommendations Made on Pinterest	117	102

a. What is the point estimate of the proportion of women who trust recommendations made on Pinterest (to 2 decimals)?

b. What is the point estimate of the proportion of men who trust recommendations made on Pinterest (to 2 decimals)?

c. Provide a 95% confidence interval estimate of the difference between the proportion of women and men who trust recommendations made on Pinterest (to 4 decimals). Use z-table.
(  to  )

Answer 1

1)

mean of sample 1,    x̅1=   104          
population std dev of sample 1,   σ1 =    8.4          
size of sample 1,    n1=   80          
                  
sample #2   --------->              
mean of sample 2,    x̅2=   106          
population std dev of sample 2,   σ2 =    7.6          
size of sample 2,    n2=   70          
                  
difference in sample means = x̅1 - x̅2 =    104   -   106   =   -2
                  
std error , SE =    √(σ1²/n1+σ2²/n2) =    1.3066          
                  
Z-statistic = ((x̅1 - x̅2)-µd)/SE =    -2   /   1.3066   =   -1.53
                  
                  
p-value =        0.1258   [excel formula =2*NORMSDIST(z)]  

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

2)

   mean of sample 1,    x̅1=   135.67          
population std dev of sample 1,   σ1 =    38          
size of sample 1,    n1=   40          
                  
sample #2   --------->              
mean of sample 2,    x̅2=   68.64          
population std dev of sample 2,   σ2 =    20          
size of sample 2,    n2=   34          
                  
point estimate of difference in sample means = x̅1 - x̅2 =    135.67   -   68.64   =   67.03
..................

b)

Level of Significance ,    α =    0.01          
z-critical value =    Z α/2 =    2.5758   [excel function =normsinv(α/2) ]      
                  
std error , SE =    √(σ1²/n1+σ2²/n2) =    6.9184          
margin of error, E = Z*SE =    2.5758   *   6.918   =   17.82
                  
c)

difference of means = x̅1 - x̅2 =    135.67   -   68.64   =   67.03
confidence interval is                   
Interval Lower Limit= (x̅1 - x̅2) - E =    67.030   -   17.821   =   49.21
Interval Upper Limit= (x̅1 - x̅2) + E =    67.030   +   17.821   =   84.85

CI (49.21 , 84.85)

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

3)

a)

female:

first sample size,     n1=   150          
number of successes, sample 1 =     x1=   117          
point estimate of proportion success of sample 1 , p̂1=   x1/n1=   0.7800          
                  
b)

sample #2   ----->   standard          
second sample size,     n2 =    170          
number of successes, sample 2 =     x2 =    102          
point estimate of proportion success of sample 1 , p̂ 2=   x2/n2 =    0.600          
                  
difference in sample proportions, p̂1 - p̂2 =     0.7800   -   0.6000   =   0.1800

c)

level of significance, α =   0.05              
Z critical value =   Z α/2 =    1.960   [excel function: =normsinv(α/2)      
                  
Std error , SE =    SQRT(p̂1 * (1 - p̂1)/n1 + p̂2 * (1-p̂2)/n2) =     0.0506          
margin of error , E = Z*SE =    1.960   *   0.0506   =   0.0991
                  
confidence interval is                   
lower limit = (p̂1 - p̂2) - E =    0.180   -   0.0991   =   0.0809
upper limit = (p̂1 - p̂2) + E =    0.180   +   0.0991   =   0.2791
                  
so, confidence interval is (   0.0809   < p1 - p2 <   0.2791   )  

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

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