Answer: 1.300228
Step-by-step explanation:
The t-score we use for the confidence interval is two-tailed , i.e. the t-score should be used to find a () is given by :-
, where n is the sample size.
Given : Level of confidence:
Significance level :
Sample size : n= 47
Then, degree of freedom :
Now by using standard normal t-distribution table,
Hence, the t-score should be used to find a 80% confidence interval estimate for the population mean = 1.300228
Answer:
Using calculator: 0.0415
Using Z-score Table: 0.0418
Step-by-step explanation:
There are two ways you can solve this problem.
1. Use the normal distribution function on a calculator.
Entered values:
Lower Limit: 126
Upper Limit: 999999999999999... (To encompass all the data)
Standard Deviation: 15
Mean: 100
2. Find the Z score and look up probabilities on table.
Formula for Z score:
Z = 1.7333
This means that the value 126 is 1.733 standard deviations away from the mean. We can look this value up on the Z table to find its corresponding probability.
This will show us the probability of the random sampling being equal to or lower than 126.
P = 0.9582
So to find the probability of it being above, we simply just calculate the inverse as all probabilities on the curve = 1.
1-0.9582 = 0.0418
NOTE: Values found from the table will usually be a bit different from if you find it from a calculator, the one you need will depend on the method you use in class.
Hope this helped!