Question

Researchers examined the reproductive characteristics of the eastern cottonmouth (a type of snake). They were able to record the number of baby snakes per litter for 24 female cottonmouths in Florida and 44 female cottonmouths in Virginia. The mean number of young per litter for the sample of cotton- mouths in Florida is 5.46 with a standard deviation of 1.59, and the mean number of young per litter for sample of Virginia cottonmouths is 7.59 with standard deviation 2.68. Preliminary data analyses indi- cate that you can reasonably presume that litter sizes of cottonmouths in both states are approximately normally distributed. Do not assume that the population standard deviations are equal.

(a) Construct a 95% confidence interval for the difference between mean litter size of Florida cottonmouths compared to Virginia cottonmouths. Write a sentence interpreting this confidence interval.

Test at the 1% significance level if the mean litter size for Florida cottonmouths is smaller than the mean litter size for Virginia cottonmouths.

(b) State the null and the alternative hypothesis using appropriate symbols.

(c) Find the value of the test-statistic and the p-value (use correct symbols).

(d) Please circle one of the following:

REJECT H0   DO NOT REJECT H0

(e) State your conclusion:

Answer 1

a)

Sample #1   ---->   1
mean of sample 1,    x̅1=   5.46
standard deviation of sample 1,   s1 =    1.59
size of sample 1,    n1=   24
      
Sample #2   ---->   2
mean of sample 2,    x̅2=   7.590
standard deviation of sample 2,   s2 =    2.68
size of sample 2,    n2=   44
      

α=0.05

Degree of freedom, DF=       65          
t-critical value =    t α/2 =    1.997   (excel formula =t.inv(α/2,df)      
                  
                  
                  
std error , SE =    √(s1²/n1+s2²/n2) =    0.518          
margin of error, E = t*SE =    1.997   *   0.518   =   1.035
                  
difference of means = x̅1-x̅2 =    5.4600   -   7.590   =   -2.1300
confidence interval is                   
Interval Lower Limit = (x̅1-x̅2) - E =    -2.1300   -   1.035   =   -3.165
Interval Upper Limit = (x̅1-x̅2) + E =    -2.1300   -   1.035   =   -1.095

b)

Ho :   µ1 - µ2 =   0
Ha :   µ1-µ2 <   0

c)

difference in sample means = x̅1-x̅2 =    5.460   -   7.5900   =   -2.1300
                  
std error , SE =    √(s1²/n1+s2²/n2) =    0.5182          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -2.1300   /   0.5182   ) =   -4.1101
                  
Degree of freedom, = 65          
                  
                  

p-value =        0.00006   [ excel function: =T.DIST(t stat,df) ]       

d)

Conclusion:     p-value<α , Reject null hypothesis  

e)at the 1% significance level there is enough evidence that the mean litter size for Florida cottonmouths is smaller than the mean litter size for Virginia cottonmouths