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 95% confidence interval to estimate the true cost of the VCR?
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A. B. C. D. E.C
Interval estimates can be found from the empirical rule where 95% will lie between plus and minus 1.96 standard deviations of the mean.
To calculate the 95% confidence interval for estimating the true cost of the VCR, we need to use the formula:
Confidence interval = Sample mean ± (Z * Standard error)
In this case, the sample mean is the average cost of the VCR in the survey, the Z-value represents the desired level of confidence (95% corresponds to a Z-value of approximately 1.96), and the standard error is calculated as the standard deviation divided by the square root of the sample size.
Given: Sample mean (x̄) = $375 Standard deviation (σ) = $20 Sample size (n) = 144 Z-value (Z) for 95% confidence = 1.96
First, let's calculate the standard error: Standard error = σ / √n Standard error = $20 / √144 Standard error ≈ $20 / 12 Standard error ≈ $1.67
Now, we can calculate the 95% confidence interval using the formula: Confidence interval = $375 ± (1.96 * $1.67)
Lower limit = $375 - (1.96 * $1.67) Lower limit ≈ $375 - $3.27 Lower limit ≈ $371.73
Upper limit = $375 + (1.96 * $1.67) Upper limit ≈ $375 + $3.27 Upper limit ≈ $378.27
Therefore, the 95% confidence interval to estimate the true cost of the VCR is approximately $371.73 to $378.27.
Among the given answer choices, the closest option is: C. $335.80 to $414.20
Please note that this answer assumes that the sample is representative and that the data follows a normal distribution.