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Answer:
a) The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).
b) The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).
c) The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).
d) The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.
Step-by-step explanation:
a) Develop a 90% confidence interval estimate of the population mean monthly rent.
Our sample size is 120.
The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So
Then, we need to subtract one by the confidence level and divide by 2. So:
Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 119 and 0.05 in the t-distribution table, we have .
Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So
Now, we multiply T and s
The lower end of the interval is the mean subtracted by M. So it is 3486 - 98.37 = $3387.63.
The upper end of the interval is the mean added to M. So it is 3486 + 98.37 = $3584.37.
The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).
b) Develop a 95% confidence interval estimate of the population mean monthly rent.
Now we have that
So
With 119 and 0.025 in the t-distribution table, we have .
The lower end of the interval is the mean subtracted by M. So it is 3486 - 117.50 = $3368.5.
The upper end of the interval is the mean added to M. So it is 3486 + 117.50 = $3603.5.
The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).
c) Develop a 99% confidence interval estimate of the population mean monthly rent.
Now we have that
So
With 119 and 0.025 in the t-distribution table, we have .
The lower end of the interval is the mean subtracted by M. So it is 3486 - 155.34 = $3330.66.
The upper end of the interval is the mean added to M. So it is 3486 + 155.34 = $3641.34.
The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).
d) What happens to the width of the confidence interval as the confidence level is increased? Does this seem reasonable? Explain.
The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.
Answer:
<h3>Answer:</h3>
- A (graph below)
- Not appropriate
- Appropriate
- Not appropriate
1st Question
The transformation g(x) = f(x -h) +k translates a function h units to the right and k units upward. Your expression has (h, k) = (-2, -4), so the graph of the function will be translated 2 units left and 4 units down.
A distinctive feature of the cube root curve is the vertical spot normally at point (0, 0). You're looking for the graph that has that relocated to (-2, -4). Graph A matches that.
2nd Question
- The method described is a reasonable one for choosing a random sample of theater patrons, but there is no reason to believe theater patrons are representative of all drivers in the city. The method is not appropriate.
- Provided the persons in the study are representative of the demographic of interest, assigning them to study groups based on the flip of a coin is appropriate.
- The variable under study is previous exposure to chicken pox, so members of the study groups need to be chosen on that basis. Participants in the study need to be carefully selected to control for all other variables. The group assignment described is not appropriate.