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>
Answer:
The null and alternative hypotheses are:
Under the null hypothesis, the test statistic is:
Where:
is the sample mean
is the sample standard deviation
is the sample size
Now, the right tailed t critical value at 0.05 significance level for df = n-1 = 10-1 = 9 is:
Since the t statistic is less than the t critical value at 0.05 significance level, therefore,we fail to reject the null hypothesis and conclude that there is not sufficient evidence to support the claim that the average phone bill has increased.
The cost will be 9 for 1 pencial
Answer:
F. (0, 5) and (-4, 2)
Step-by-step explanation:
We need to calculate the slope of each of the given sets of points until we find the set associated with a slope of 3/4:
F. (0,5) and (-4,2) As we go from (-4, 2) to (0,5), x increases by 4 and y increases by 3, so the slope is m = rise / run = 3/4. This is the line with slope 3/4.
34m + 14 = 30m + 22
34m - 30m = 22 - 14
4m = 8
m = 8/4
m = 2...multiple choice questions are worth 2 points each
