Assume that the national credit card interest rate is 12.83 percent. A study of 62 college students finds that their average int
erest rate is 14 percent with a standard deviation of 9.3 percent. What is the 95 percent confidence interval for this single-sample t test?
a. [6.94, 11.66]
b. [11.64, 16.36]
c. [10.47, 15.19]
d. [12.99, 15.01]
a. [6.94, 11.66]
b. [11.64, 16.36]
c. [10.47, 15.19]
d. [12.99, 15.01]
1 answer:
Answer:
b) 95 percent confidence interval for this single-sample t test
[11.64, 16.36]
Step-by-step explanation:
Explanation:-
Given data a study of 62 college students finds that their average interest rate is 14 percent with a standard deviation of 9.3 percent.
Sample size 'n' =62
sample mean x⁻ = 14
sample standard deviation 'S' = 9.3
<u>95 percent confidence interval for this single-sample t test</u>
The values are the <u>95 percent confidence interval for the population mean 'μ'</u>
Degrees of freedom γ=n-1=62-1=61
t₀.₀₅ = 1.9996 at 61 degrees of freedom
(14-2.361 , 14 + 2.361)
[(11.64 , 16.36]
<u>Conclusion:-</u>
95 percent confidence interval for this single-sample t test
[11.64, 16.36]
<u></u>
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32 × 2 = 68
32 × 3 = 96
32 × 4 = 128
So far we can conclude that the answer to your question lies somewhere between the numbers 3 and 4. To narrow down the answer some more, multiply 32 by 3.5 (a midway point between 3 and 4).
32 × 3.5 = 112
The number 112 tells us that the decimal we are looking for is higher than 3.5. (Because we need to get to 125, not 112.) Let's try some decimals between 3.5 and 4.
32 × 3.7 = 118.4
32 × 3.8 = 121.6
32 × 3.9 = 124.8
32 × 4 = 128
As we narrow down our answer, we can see that the number we are looking for lies between 3.9 and 4 on the number line. Now we need to start testing some decimals between 3.9 and 4.
32 × 3.905 = 124.96
Again, use the number five as a "midway" point to decide if you should use numbers that are higher or lower than 3.905. In this case, we need to use numbers higher than 3.905.
32 × 3.906 = 124.992
32 × 3.907 = 125.024
We are getting even closer to our number now that we know the decimal is somewhere between 3.906 and 3.907.
32 × 3.9065 = 125.008
With our midway point we can see that our number lies between 3.906 and 3.9065. Let's try a quarter point to see where our number lies from there.
32 × 3.90625 = 125
And BINGO! We have found the answer to the question. To be rephrased, our answer can be put like this:
= 3.90625
32 × 3 = 96
32 × 4 = 128
So far we can conclude that the answer to your question lies somewhere between the numbers 3 and 4. To narrow down the answer some more, multiply 32 by 3.5 (a midway point between 3 and 4).
32 × 3.5 = 112
The number 112 tells us that the decimal we are looking for is higher than 3.5. (Because we need to get to 125, not 112.) Let's try some decimals between 3.5 and 4.
32 × 3.7 = 118.4
32 × 3.8 = 121.6
32 × 3.9 = 124.8
32 × 4 = 128
As we narrow down our answer, we can see that the number we are looking for lies between 3.9 and 4 on the number line. Now we need to start testing some decimals between 3.9 and 4.
32 × 3.905 = 124.96
Again, use the number five as a "midway" point to decide if you should use numbers that are higher or lower than 3.905. In this case, we need to use numbers higher than 3.905.
32 × 3.906 = 124.992
32 × 3.907 = 125.024
We are getting even closer to our number now that we know the decimal is somewhere between 3.906 and 3.907.
32 × 3.9065 = 125.008
With our midway point we can see that our number lies between 3.906 and 3.9065. Let's try a quarter point to see where our number lies from there.
32 × 3.90625 = 125
And BINGO! We have found the answer to the question. To be rephrased, our answer can be put like this: