A survey of 144 retail stores revealed that a particular brand and model of a VCR retails for $375 with a standard deviation of $20. What is the 99% confidence interval to estimate the true cost of the VCR?
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A. B. C. D. E.B
For 99%, the z value will be 2.575. Therefore we have 375 +/- 2.575*20.
To calculate the 99% confidence interval for estimating the true cost of the VCR, we'll use the formula:
Confidence Interval = Sample Mean ± (Z * Standard Error)
Where: Sample Mean: The average cost of the VCR in the sample. Z: The Z-score corresponding to the desired confidence level. Standard Error: The standard deviation of the sample divided by the square root of the sample size.
First, let's find the sample mean. The survey revealed that the retail price of the VCR is $375. Therefore, the sample mean is $375.
Next, we need to find the Z-score. The Z-score corresponds to the desired confidence level, which in this case is 99%. Since it's a two-tailed test (we're looking for the interval), we'll divide the desired confidence level by 2 to find the area in each tail. From the standard normal distribution table or calculator, we find that the area in each tail is 0.005. Subtracting this value from 1 gives us the area in the middle, which is 0.99. The Z-score corresponding to this area is approximately 2.576.
Now, let's calculate the standard error. The standard deviation given is $20, and the sample size is 144. So, the standard error is calculated as follows:
Standard Error = Standard Deviation / √(Sample Size) = $20 / √144 = $20 / 12 = $1.67 (rounded to two decimal places)
Finally, we can calculate the confidence interval:
Confidence Interval = Sample Mean ± (Z * Standard Error) = $375 ± (2.576 * $1.67) = $375 ± $4.29 = ($375 - $4.29, $375 + $4.29) = ($370.71, $379.29)
Therefore, the 99% confidence interval to estimate the true cost of the VCR is approximately $370.71 to $379.29.
Among the given answer choices, none of them match the calculated confidence interval.