Question

SOLVE 1-3

1. During the last three presidential elections, I claim that the mean age of a voter was greater than 30 years of age.

2. During the last presidential election, I believe that at least 50 percent of female voters (ages 18-49) voted for Hillary Clinton.

3. Dave likes to grow cucumbers, and each year he experiments with ways to improve his crop. This summer, he wants to determine whether the new fertilizer he has seen advertised will increase the mean number of cucumbers produced per plant. He uses the old fertilizer on 15 cucumber plants and the new fertilizer on the other 10 plants. Over the course of the growing season, he calculates a mean of 17.4 cucumbers per plant for the old fertilizer, with a standard deviation of 1.8 tomatoes. He calculates a mean of 20.4 cucumbers per plant for the new fertilizer, with a standard deviation of 2.7 tomatoes. You may assume that both samples came from populations that are approximately normal. Assume that the population variances are not equal

(a) Determine the best point estimate for the true mean difference in cucumber production. Correct notation counts

(b) Construct a 95% confidence interval for the true mean difference in cucumber production. For credit, your work must include a filled-out formula for your confidence interval

(c) Interpret the confidence interval in the context of the original problem.

Answer 1

a)

mean of old,    x̅1=   17.40                  
standard deviation of sample 1,   s1 =    1.80                  
size of sample Old,    n1=   15                  
                          
Sample New ---->   New   
mean of sample New,    x̅2=   20.40                  
standard deviation of sample 2,   s2 =    2.70                  
size of sample New,    n2=   10                  
                          
difference in sample means =    x̅1-x̅2 =    17.4000   -   20.4   =   -3.00  

b)

Degree of freedom, DF =    14              
t-critical value =    t α/2 =    2.1448   (excel formula =t.inv(α/2,df)          
                            
                      
std error , SE =    Sp*√(1/n1+1/n2) =    0.9721              
margin of error, E = t*SE =    2.1448   *   0.9721   =   2.0850  
                      
difference of means =    x̅1-x̅2 =    17.4000   -   20.400   =   -3.0000
confidence interval is                       

Interval Lower Limit=   (x̅1-x̅2) - E =    -3.0000   -   2.0850   =   -5.0850
Interval Upper Limit=   (x̅1-x̅2) + E =    -3.0000   +   2.0850   =   -0.9150

c)

As we can see no zero in the confidence interval we can say that...............
The new fertilizer has seen increase in the mean number of cucumbers produced per plant vs old fertiliser.

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