Question

18)

1-

A researcher samples 15 individuals from a population of butterflies and measures their body mass at pupation, and wants to compare the sample to a previously-reported mean of 1.2g using a one-sample t-test. The sample mean is 0.97g with a standard error of 0.14. What is the value of the t-statistic?

Select one:

a. -1.643

b. 1.643

c. 0.123

d. 0.231

2-

You have 15 coins in your pocket and select 10 at random. If you do not know the values of the coins, which of the following is TRUE regarding the number of possible outcomes?

Select one:

a. There are fewer outcomes than if you select 7 at random.

b. There are fewer outcomes than if you select 5 at random.

c. There are fewer outcomes than if you select 3 at random.

d. There are fewer outcomes than if you select 1 at random.

3-

Normally a null hypothesis is phrased in the negative (e.g., “the means are not different”) because:

Select one:

a.
We cannot prove a hypothesis is correct, but we can prove a hypothesis is incorrect.

b.
It’s arbitrary.

c.
This is a requirement of the statistical tests.

d.
R.A. Fisher first used this format in dummy experiments and it has stuck.

4-

You have two independent samples, each with 50 observations, and you want to compare their means. The mean of the first sample is 6.5 and the variance is 1.5. The mean of the other is 5.25 and the variance is 1.4. What is the value of t for a Student’s t-test (rounded to 3 decimal places)?

Select one:

a. 7.245

b. -21.740

c. -15.299

d. 5.190

Answer 1

18)

µ =   1.2

Sample Size ,   n =    15
Sample Mean,    x̅ =   0.9700
            
Standard Error , SE = s/√n =   0/√15=   0.1400
t-test statistic= (x̅ - µ )/SE =    (0.97-1.2)/0.14=   -1.643 (a)

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

3)

a) We cannot prove a hypothesis is correct, but we can prove a hypothesis is incorrect.

...........

4)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   1                  
mean of sample 1,    x̅1=   6.50                  
standard deviation of sample 1,   s1 =    1.22                  
size of sample 1,    n1=   50                  
                          
Sample #2   ---->   2                  
mean of sample 2,    x̅2=   5.25                  
standard deviation of sample 2,   s2 =    1.18                  
size of sample 2,    n2=   50                  
                          
difference in sample means =    x̅1-x̅2 =    6.5000   -   5.3   =   1.25  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.2042                  
std error , SE =    Sp*√(1/n1+1/n2) =    0.2408                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   1.2500   -   0   ) /    0.24   =   5.190

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

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