When the population standard deviation (σ) is not known, why is the sample standard deviation (s >) used to calculate probability?
A)
There is only one possible value for the population standard deviation for a given population, . How many possible values are there for the sample standard deviation taken from this population?
A)
Why is the t-Curve wider than the z-Curve?
D)
The t-Distribution cannot be used to find the probability of an event.
True
False
False
What is the shape of the t-Distribution?
D)
What is the location of the t-Distribution?
A)
What is the spread of the t-Distribution?
C)
t-Values are denoted: t(α, n-1)
Please match the following symbols with their meaning.
A- t
B- n-1
C- α
__ A point on the x-axis
__ The area in the tail (right or left) of the curve
__ The degrees of freedom
A
C
B
t-Values are denoted: t(α, n-1)
Please match the following symbols with where their value is found in the t-table.
A- n-1 (degrees of freedom)
B- α (area in one tail)
C- t (t-t-vlues)
__ In the left column of the t-table
__ In the body of the t-table
__ In the top row of the t-table
A
C
B
What is the first step in using the t-Table?
A)
A new workout app claims that users could burn 350 calories on average per workout session. A local statistics student questions this claim so she gets 12 of her friends to follow the workout app and calculates the average number of calories burned. The standard deviation of this sample of calories burned was found to be 25 calories. What is the probability that the sample average of her friends is over 363?
A) 0.05
B) 1.801
C) 0.10
D) 2.201
t= (x-μ) / (s / √n)
= (363 - 350) / (25/√12)
=1.801
Using t-table: df= 12-1 = 11
Go to 11 df on the left and go horizontal until you find 1.801. The number on the upper x-axis is the answer.
= 0.005
A)
A new workout app claims that users could burn 350 calories on average per workout session. A local statistics student questions this claim so she gets 12 of her friends to follow the workout app and calculates the average number of calories burned. The standard deviation of this sample of calories burned was found to be 25 calories. What is the probability that the sample average of her friends is less than 340 or greater than 360 calories burned?
A) 0.20
B) 0.10
C) 0.40
D) 0.14
Since it is sample population, we use t-chart.
t= (x-μ) / (s / √12)
= (340 - 350) / (25 / √12) = 0.10
=(360 - 350) / (25 / √12) = 0.10
0.10+0.10 = 0.20
A)
A kindergarten teacher had an incoming class of 28 students that seemed to be both shorter and taller than usual. She measured these students, calculated the sample average height, and calculated a standard deviation height of 2.6 inches. If the population mean height of incoming kindergarten students was known to be 39 inches, what is the probability of the average height of this class being less than 37.5 inches or greater than 40.5 inches?
A) 0.005
B) 0.0025
C) 0.01
D) 0.02
t= (x-μ) / (s / √n)
= (37.5 - 39) / (2.6 / √28) = -3.05
= (40.5 - 39) / (2.6 / √28) = +3.05
df= 28-1 = 27
Using t-chart, df=27 and t-value= 3.05, we get tail length of 0.0025 for both
0.0025 + 0.0025 = 0.0050
A)
Use the t-table to find the correct t-value for a sample size of 6 and a right tail area of 0.10.
Use t-table
Sample size 6 = 1 less df
So, 6-1 = 5 df
5df and tail area of 0.10= 1.476
Use the t-table to find the correct t-value for a sample size of 18 and a left tail area of 0.05.
sample size= 18 == df = 17
df= 17 and tail area of 0.05= 1.740
Use the t-table to find the correct tail area for a sample size of 20 and a t-value 2.539.
sample size= 20, so df= 19
df= 19 and t-value= 2.549 = 0.01
A town in Sweden was known to have long-lived residents with a mean life span of 80 years. To find out if this was still true, a sociology researcher looked at several samples of 40 residents that had a standard deviation life span of 8 years. What would be the lowest and the highest life span of the middle 90% these residents?
A) Between 76.7 years and 83.3 years.
B) Between 77.9 years and 82.1 years.
C) Between 79.2 years and 80.7 years.
D) Between 79.7 years and 80.3 years.
The middle section is 90% meaning the tail areas are 5% each.
df= 40-1 = 39
0.05 tail areas
So, using t-table, if we go to df=39 and 0.05 tail area, we get: t= 1.685
t= (x-μ) / (σ / √n)
+1.685= (x-80) / (8/√40) == 82.1
-1.685= (x-80) / (8/√40) == 77.9
B)
A new workout app claims that users could burn 350 calories on average per workout session. A local statistics student questions this claim so she gets 12 of her friends to follow the workout app and calculates the average number of calories burned. The standard deviation of this sample of calories burned was found to be 25 calories. What is the probability that the sample average of her friends is less than 340 or greater than 360 calories burned?
A) 0.10
B) 0.20
C) 0.40
D) 0.14
t= (x-μ) / (σ / √n)
= (360-350) / (25 / √12) == +1.38
= (340-350) / (25 / √12) == -1.38
df= 12-1 = 11
Using df=11 and going to the closes value to 1.38, we get tail area of 0.10.
Since both tails are symmetrical, we can add those values together, so 0.10+2= 0.20
B)
How to calculate standard error
standard deviation / √n
A fishery graduate student went fishing in the university lake where the fish were known to be 14 inches long on average. The graduate student caught 8 fish and calculated a standard deviation length of 1.5 inches. What would he expect to be the shortest and the longest average length of the middle 99% of samples?
A) Between 12.92 and 15.07 inches long.
B) Between 13.34 and 14.65 inches long.
C) Between 11.15 and 16.85 inches long.
D) Between 12.14 and 15.85 inches long.
Since the middle section is 99% of the area, the tails consist of 1% (0.5% each). 0.5% = 0.005
