An internet supplier of refilled ink cartridges for ink jet printers sold cartridges to 30,000 customers over the past two months. A random sample of 1000 of those customers revealed that 18% were not happy with their purchase. The margin of error was 2%. What range is likely to contain the population parameter?
29,000 to 31,000
14% to 22%
16% to 20%
480 to 600
.18 + or - .02 = range likely to contain population parameter
.18 + .02 = .20
.18 - .02 = .16
16% to 20% -->converting decimals to percents
To determine the range that is likely to contain the population parameter, we can calculate the margin of error and then create a confidence interval.
The margin of error formula is given by:
Margin of Error = Critical Value * Standard Error
Since the sample size is large (n = 1000) and the statistic being measured (not happy with the purchase) is a proportion, we can assume that the sampling distribution follows a normal distribution. In this case, we can use the standard normal distribution, with a critical value corresponding to the desired level of confidence. If we assume a 95% confidence level, the critical value is approximately 1.96.
The standard error formula for proportions is given by:
Standard Error = sqrt((p * (1 - p)) / n)
In this case, the sample proportion (p) is 18% (0.18) and the sample size (n) is 1000.
Now, let's calculate the margin of error:
Margin of Error = 1.96 * sqrt((0.18 * (1 - 0.18)) / 1000)
Margin of Error ≈ 1.96 * sqrt(0.14688 / 1000)
Margin of Error ≈ 1.96 * 0.0121
Margin of Error ≈ 0.0237
The margin of error is approximately 0.0237.
To create the confidence interval, we subtract and add the margin of error to the sample proportion:
Lower Bound = 0.18 - 0.0237 = 0.1563
Upper Bound = 0.18 + 0.0237 = 0.2037
Therefore, the range that is likely to contain the population parameter is approximately 15.63% to 20.37%.
Among the options provided, the answer is:
16% to 20%