Determine A 95% Confidence Interval For The Proportion Defec

Determine a 95% confidence interval for the proportion defective for the process today

The percent defective for parts produced by a manufacturing process is targeted at 4%. The process is monitored daily by taking samples of size n = 160 units. Today’s sample contains 14 defectives. Determine a 95% confidence interval for the proportion defective for the process today. Place your LOWER limit, rounded to 3 decimal places, in the first blank. Place your UPPER limit, rounded to 3 decimal places, in the second blank.

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The task involves calculating a 95% confidence interval for the proportion of defective units in a manufacturing process, based on a sample. Given that the process aims for a defect rate of 4%, with a sample size of 160 units, and 14 defectives observed, the estimator for the proportion defective (p̂) is calculated as:

p̂ = number of defectives / sample size = 14 / 160 = 0.0875.

The confidence interval for a population proportion is calculated using the formula:

p̂ ± Z_α/2 * √[p̂(1 - p̂) / n],

where Z_α/2 is the critical value from the standard normal distribution for a 95% confidence level, which is approximately 1.96.

Calculating the standard error (SE):

SE = √[0.0875 (1 - 0.0875) / 160] = √[0.0875 0.9125 / 160] = √[0.0799 / 160] ≈ √0.000499 ≈ 0.0223.

Multiplying by Z-value:

Margin of Error (ME) = 1.96 * 0.0223 ≈ 0.0437.

Now, the 95% confidence interval is:

Lower limit: 0.0875 - 0.0437 ≈ 0.0438.

Upper limit: 0.0875 + 0.0437 ≈ 0.1312.

Rounded to three decimal places, the confidence interval for the proportion defective today is:

Lower limit: 0.044

Upper limit: 0.131

This interval suggests that, with 95% confidence, the true proportion of defective units for today’s process lies between approximately 4.4% and 13.1%, reflecting uncertainty around the point estimate of 8.75% defect rate observed.

Estimate Sample Size for Estimating Mean Salary Within a Margin of Error

Given that a sample of 9 production managers shows an average salary of $71,000 with a standard deviation of $18,000, and the goal is to estimate the true mean salary within $4,200 with 95% confidence, the required sample size n can be determined by the formula:

n = (Z_α/2 * s / E)^2.

Here, Z_α/2 for 95% confidence is approximately 1.96, s = 18000, and the margin of error E = 4200.

Calculating n:

n = (1.96 18000 / 4200)^2 = (1.96 4.2857)^2 ≈ (8.399)^2 ≈ 70.55.

Since the sample size must be a whole number, we round up to get:

n = 71.

Therefore, approximately 71 managers need to be sampled for the estimate.

Constructing Confidence Interval for Mean Salary

Assuming that salaries are normally distributed, with a sample mean of $71,000 and standard deviation of $18,000 based on 9 managers, to find a 95% confidence interval for the mean salary, the t-distribution is used due to small sample size (n=9). The degrees of freedom: df = n - 1 = 8.

The t-critical value for 8 degrees of freedom at 95% confidence is approximately 2.306.

The standard error (SE) is:

SE = s / √n = 18000 / √9 = 18000 / 3 = 6000.

The margin of error (ME):

ME = 2.306 * 6000 ≈ 13,836.

The confidence interval bounds are:

Lower limit: 71,000 - 13,836 ≈ 57,164.

Upper limit: 71,000 + 13,836 ≈ 84,836.

Thus, the 95% confidence interval for the mean salary of all managers with over 15 years of experience is approximately between $57,164 and $84,836.

Sample Size for Estimating Population Proportion with a Higher Confidence and Smaller Error

Initially, a sample size of 1000 was used to estimate a proportion. To reduce the maximum allowable error from 0.05 to 0.025, the new required sample size n' can be calculated using the relation:

n' = n * (E_old / E_new)²

Given n = 1000, E_old = 0.05, E_new = 0.025:

n' = 1000 (0.05 / 0.025)^2 = 1000 (2)^2 = 1000 * 4 = 4000.

Therefore, the new sample size should be 4000.

Determining Sample Size for Estimating Average Family Food Expenditure

To estimate average family food expenditure within $60, given a standard deviation of $1200, and a 99% confidence level where Z_α/2 ≈ 2.576, the required sample size is:

n = (Z_α/2 σ / E)^2 = (2.576 1200 / 60)^2 = (2.576 * 20)^2 ≈ (51.52)^2 ≈ 2656.41.

Rounding up, about 2657 families need to be surveyed.

Confidence Interval for Variance of Life Expectancy

A sample of 23 European countries reveals a variance in life expectancy of 7.3 years. The 95% confidence interval for the population variance σ² is constructed using the Chi-square distribution:

Lower bound: ((n - 1) * s²) / χ²_upper

Upper bound: ((n - 1) * s²) / χ²_lower

Degrees of freedom: df = 22.

From Chi-square tables, χ²_{0.025, 22} ≈ 10.980, and χ²_{0.975, 22} ≈ 36.780.

Calculations:

Lower limit: (22 * 7.3) / 36.780 ≈ 160.6 / 36.780 ≈ 4.37.

Upper limit: (22 * 7.3) / 10.980 ≈ 160.6 / 10.980 ≈ 14.63.

Since the variance is 7.3, the confidence interval bounds for variance are approximately between 4.4 and 14.6 years².

Sample Size for Estimating Mean Nickel Weight

With a standard deviation of 150 mg, to be 98% confident that the sample mean is within 25 mg of the true mean, the required sample size:

n = (Z_α/2 * σ / E)^2.

For 98% confidence, Z_α/2 ≈ 2.33.

Calculations:

n = (2.33 150 / 25)^2 = (2.33 6)^2 ≈ (13.98)^2 ≈ 195.44.

Rounding up, at least 196 nickels should be weighed.

Probability That a Randomly Selected Employee Has Worked Over 10 Years

Given average tenure of 5.7 years with standard deviation 1.8 years, and assuming normality, the probability that a randomly selected employee has worked more than 10 years is calculated via the Z-score:

Z = (X - μ) / σ = (10 - 5.7) / 1.8 ≈ 4.3 / 1.8 ≈ 2.39.

The probability that Z > 2.39 is 1 - Φ(2.39). From standard normal tables, Φ(2.39) ≈ 0.9916.

Thus, the probability P(X > 10) ≈ 1 - 0.9916 = 0.0084, or about 0.84%.

Using t-Distribution for 15 Degrees of Freedom

The value of P(T 15) corresponds to the cumulative probability up to the t-value with 15 degrees of freedom. For example, if t₁₅ = 0.03197, the probability is approximately 0.03197, and similarly for the others. The exact value depends on the specific t-value queried.

Score Needed to Be in the Top 5% of SAT Scores

Given a mean of 1000 and a standard deviation of 200, the lowest score to be in the top 5% corresponds to the 95th percentile. The Z-score for the 95th percentile is approximately 1.645. The cutoff score:

X = μ + Z σ = 1000 + 1.645 200 = 1000 + 329 ≈ 1329.

Rounded to the nearest whole number, the minimum score needed is approximately 1330.

Confidence Interval for Proportion of Women Internet Users in Europe

Out of 750 internet users, 35% (262.5, rounded to 263) are women. The 95% confidence interval for the true proportion p is:

p̂ ± Z_α/2 * √[p̂(1 - p̂) / n].

p̂ = 0.35, Z_0.975 ≈ 1.96.

Calculating standard error:

SE = √[0.35 * 0.65 / 750] ≈ √[0.2275 / 750] ≈ √0.000303 ≈ 0.0174.

Margin of error: 1.96 * 0.0174 ≈ 0.0341.

Lower limit: 0.35 - 0.0341 ≈ 0.3159.

Upper limit: 0.35 + 0.0341 ≈ 0.3841.

Therefore, the 95% confidence interval is approximately 0.316

Additional Notes

Discussions on the appropriateness of normal or t-distribution approximations depend on the sample size and data distribution. For sample proportions with n > 30, normal approximation is generally acceptable; for smaller samples or more precise estimates, t or chi-square distributions are used accordingly.

References

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