Given the information: the sampled population is normally distributed, = 10.4, n = 60, and = 81.2.
a. Find the 98% confidence interval for
b. Interpret the confidence interval in
Where do i begin
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In order to monitor the quality of a production process, samples of size five are selected daily. Thee random variable of interest is the number of defectives in the five items selected. What values are possible for this random variable? Solutions
To begin, you will need to calculate the confidence interval for the given information. The formula for calculating a confidence interval for a population mean when the population standard deviation is known is:
Confidence Interval = (sample mean ± (critical value * standard deviation / √n))
Let's break down the steps:
a. Find the 98% confidence interval:
1. Identify the given data:
- Mean (μ): 10.4
- Sample Size (n): 60
- Standard Deviation (σ): 81.2
2. Determine the critical value:
The critical value corresponds to the level of confidence and is based on the desired level of significance alpha (α), which is obtained from a standard normal distribution table or a statistical calculator. For a 98% confidence level, the alpha value is (1 - 0.98) / 2 = 0.01 (since we're looking for the middle 98%).
By referring to the standard normal distribution table, the z-score corresponding to a cumulative probability of 0.99 is approximately 2.33.
3. Calculate the confidence interval:
Using the formula mentioned earlier, substitute the values into the equation:
Confidence Interval = (10.4 ± (2.33 * 81.2 / √60))
This becomes:
Confidence Interval = (10.4 ± 19.55)
Simplifying further:
Lower Limit: 10.4 - 19.55 = -9.15
Upper Limit: 10.4 + 19.55 = 29.95
Thus, the 98% confidence interval is (-9.15, 29.95).
b. Interpret the confidence interval:
The 98% confidence interval means that if we were to generate multiple samples from the same population and calculate a 98% confidence interval for each sample, then approximately 98% of these intervals would contain the true population mean.
In this specific case, we are 98% confident that the true population mean lies within the interval (-9.15, 29.95). This interval provides an estimate of the range in which the population mean is likely to fall.