In order to make sure that their bags of chips are of sufficient weight, the Chipper Chip Company try to fill their 200 gram bags of chips with more than 203 grams. You have been given the task of investgating what proportion of bags produced do have over 203 grams of chips in them. you sample 112 bags and find that the proportion of these with over 203 grams is 0.77. You decide to construct a 95% confidence interval for the population proportion of chip bags that have more than 203 grams of chips.

find the upper and lower bounds for confidence interval...

i know the steps involved to complete this question but im not sure if the mean is 200 or 203... thanks for your help

In this case, the mean is not 200 or 203 because we are not dealing with a continuous variable (weight of chips). We are interested in the proportion of bags that have more than 203 grams of chips.

To construct a confidence interval for a proportion, we can use the formula:

CI = p̂ ± Z * √((p̂ * (1 - p̂)) / n)

Where:
- p̂ is the sample proportion (0.77 in this case)
- Z is the z-score corresponding to the desired confidence level (95% confidence level corresponds to a Z-score of approximately 1.96)
- n is the sample size (112 bags)

Now, we can plug in the values:

CI = 0.77 ± 1.96 * √((0.77 * (1 - 0.77)) / 112)

Calculating the confidence interval:

CI = 0.77 ± 1.96 * √(0.177 / 112)

CI = 0.77 ± 1.96 * 0.042

CI = 0.77 ± 0.082

Therefore, the lower bound of the confidence interval is 0.77 - 0.082 = 0.688, and the upper bound is 0.77 + 0.082 = 0.852.

So, the 95% confidence interval for the proportion of chip bags that have more than 203 grams of chips is approximately 0.688 to 0.852.

To construct a confidence interval for the population proportion, we need to determine the point estimate and the appropriate critical value.

In this case, the point estimate is the proportion of bags in the sample that contain more than 203 grams of chips, which is 0.77.

The mean (μ) in this context would represent the average weight of the bags, which is not relevant for constructing a confidence interval for proportion.

Next, we need to determine the critical value using a standard normal distribution. Since the sample size is large (112 bags) and we want a 95% confidence interval, we can use the standard normal distribution (Z-distribution) with a Z-value corresponding to a 95% confidence level.

The critical value corresponding to a 95% confidence level is approximately 1.96 (obtained from a Z-table or calculator).

Now, we can calculate the upper and lower bounds for the confidence interval:

Step 1: Calculate the margin of error.
Margin of Error = Critical Value * Standard Error
Standard Error = sqrt[(p * (1-p)) / n]
where p is the sample proportion and n is the sample size.

Margin of Error = 1.96 * sqrt[(0.77 * (1-0.77)) / 112]
Margin of Error ≈ 0.084

Step 2: Calculate the upper and lower bounds.
Lower Bound = Sample Proportion - Margin of Error
Lower Bound = 0.77 - 0.084 ≈ 0.686

Upper Bound = Sample Proportion + Margin of Error
Upper Bound = 0.77 + 0.084 ≈ 0.854

Therefore, the 95% confidence interval for the proportion of chip bags with more than 203 grams of chips is approximately 0.686 to 0.854.