A random sample of 30 students spent an average amount of $40.00 on a date, with a standard deviation of $5.00. Set confidence level at 92%. a) Calculate the standard error. b) Cal... A random sample of 30 students spent an average amount of $40.00 on a date, with a standard deviation of $5.00. Set confidence level at 92%. a) Calculate the standard error. b) Calculate the margin of error. c) Calculate an approximate 92% confidence interval for the average amount spent by all students on a date. Interpret. d) If we want to change the margin of error to ±$2.00, what sample size is needed?
Understand the Problem
The question is asking to calculate various statistical measures based on a sample of students regarding the amount they spent on dates. It requires finding the standard error, margin of error, a confidence interval, and determining the sample size needed for a specific margin of error.
Answer
a) $0.91$ b) $1.60$ c) $(38.40, 41.60)$ d) $61$
Answer for screen readers
a) The standard error is approximately $0.91$.
b) The margin of error is approximately $1.60$.
c) The 92% confidence interval is approximately $(38.40, 41.60)$.
d) A sample size of approximately 61 is needed for a margin of error of ±$2.00$.
Steps to Solve
- Calculate the Standard Error
The standard error (SE) is calculated using the formula:
$$ SE = \frac{s}{\sqrt{n}} $$
where ( s ) is the standard deviation and ( n ) is the sample size.
In this case, ( s = 5.00 ) and ( n = 30 ).
So,
$$ SE = \frac{5.00}{\sqrt{30}} $$
- Calculate the Margin of Error
The margin of error (ME) at a given confidence level is calculated using:
$$ ME = z \cdot SE $$
where ( z ) is the z-score corresponding to the confidence level. For a 92% confidence level, the z-score is approximately 1.75.
Now substitute the previously calculated SE from step 1 into the formula:
$$ ME = 1.75 \cdot SE $$
- Calculate the Confidence Interval
The confidence interval (CI) is calculated using:
$$ CI = \bar{x} \pm ME $$
where ( \bar{x} ) is the sample mean. Here, ( \bar{x} = 40.00 ).
So, substitute the mean and the ME from step 2:
$$ CI = 40.00 \pm ME $$
- Calculate the Required Sample Size for a Specific Margin of Error
To find the sample size (( n )) needed for a specific margin of error, use the formula:
$$ n = \left( \frac{z \cdot s}{ME} \right)^2 $$
In this case, set ( ME = 2.00 ) and use the z-score from step 2. Substitute the values into the formula:
$$ n = \left( \frac{1.75 \cdot 5.00}{2.00} \right)^2 $$
a) The standard error is approximately $0.91$.
b) The margin of error is approximately $1.60$.
c) The 92% confidence interval is approximately $(38.40, 41.60)$.
d) A sample size of approximately 61 is needed for a margin of error of ±$2.00$.
More Information
The standard error measures how much the sample mean is expected to vary from the true population mean. The margin of error helps in understanding how confident we can be that the population parameter lies within the confidence interval. The confidence interval gives a range within which we believe the true average amount spent lies, based on our sample. Lastly, calculating the required sample size ensures the desired precision in our estimates.
Tips
- Confusing the standard deviation and standard error; recall that standard error accounts for sample size.
- Incorrectly interpreting z-scores; double-check that you are using the correct table or calculator for confidence levels.
- Not rounding properly; be consistent with significant figures, especially in final answers.
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