Sample Of 135 Results In 108 Successes Use Table 1 To Calcul

Sample Of 135 Results In 108 Successes Use Table 1acalculate Th

Determine the point estimate for the population proportion of successes based on a sample of 135 results with 108 successes, and construct confidence intervals at 95% and 99% levels to assess whether the population proportion differs from 0.860. Use the confidence intervals to evaluate the hypotheses at these levels. Additionally, analyze data related to college student debt, stock prices, customer visits, quality sampling, and other metrics, applying appropriate statistical methods such as probability calculations, confidence intervals, and hypothesis testing, with all calculations to be rounded and formatted as specified.

Paper For Above instruction

The analysis begins with estimating the population proportion of successes utilizing the sample data of 135 results, where 108 are successes. The point estimate of the population proportion (p̂) is calculated as the ratio of successes to total results —p̂ = 108/135 ≈ 0.800. This estimate provides the basis for constructing confidence intervals and testing hypotheses about the true population proportion.

To construct confidence intervals, we apply the formula for a confidence interval for a proportion: p̂ ± Z√(p̂(1 - p̂)/n), where Z is the critical value corresponding to the desired confidence level, p̂ is the point estimate, and n is the sample size. For 95% confidence, with Z ≈ 1.96, the standard error (SE) is √(0.800×0.200/135) ≈ 0.0310, leading to an interval of 0.800 ± 1.96×0.0310, which results in (0.739, 0.861). For 99% confidence, with Z ≈ 2.58, the interval is 0.800 ± 2.58×0.0310, or (0.720, 0.880). These intervals allow us to test whether the population proportion likely differs from 0.860.

At the 99% confidence level, since the interval (0.720, 0.880) contains 0.860, there is insufficient evidence to conclude the proportion differs from 0.860. Conversely, at 95%, because the interval (0.739, 0.861) nearly includes 0.860 but the upper bound exceeds it slightly, the evidence is still inconclusive for a difference at this level. The interpretation hinges on whether 0.860 is within these bounds, which indicates the plausibility of the population proportion.

Further analysis involves assessing the probability that the average debt among recent college graduates exceeds specified thresholds, assuming normality. With a mean debt of $27,400 and a standard deviation of $5,000, the probability that the average debt of three graduates exceeds $20,000 is calculated via the Z-score: Z = (20,000 - 27,400)/(5,000/√3) ≈ -4.04. Using standard normal distribution tables, the probability that the average debt exceeds $20,000 is approximately 1.0. For the $25,000 threshold, Z ≈ (25,000 - 27,400)/(5,000/√3) ≈ -1.66, yielding a probability of approximately 4.8%. These calculations help understand the likelihood of different debt outcomes among groups of graduates.

Analyzing stock prices, the sample mean and standard deviation of the monthly closing prices over six months are computed. For example, with prices of $5, $6, $9, $8, $4, and the last months, the sample calculations (mean ≈ $6.33, standard deviation ≈ $2.36) inform the construction of a 90% confidence interval for the mean stock price. Using the t-distribution with degrees of freedom = 5 and t-value ≈ 2.015, the confidence interval is (approximately $4.4, $8.2). As confidence levels increase, the margin of error expands, making the interval wider, which reduces precision but increases confidence.

Customer visit data collected over weekdays are used to form confidence intervals for the mean number of daily visitors. For the 7 observations (83, 46, 38, 34, 47, 42, 48), the sample mean is approximately 44.29 and the sample standard deviation about 14.87. The 90% confidence interval, using t-value ≈ 2.447, ranges from roughly 36.4 to 52.2. For higher confidence at 99%, with t ≈ 3.499, the interval widens further, reflecting the principle that increasing confidence level broadens the interval, decreasing the precision of the estimate.

Sample size calculations for a cereal packaging machine aim to achieve a margin of error of at most 0.01 pounds with 99% confidence. Using the formula n = (Zσ / E)^2, with Z ≈ 2.576, σ = 0.06, and E = 0.01, yields a required sample size of approximately 3850. Larger sample sizes are necessary for higher confidence and smaller errors, exemplifying the trade-off between sample size, confidence, and precision.

Regarding quality control via acceptance sampling, the probability of adjusting production machines when the proportion of nonconforming items exceeds 16% is calculated using the binomial distribution approximated by a normal distribution. For a batch of 56 items with a known nonconformance rate of 12%, the probability that the proportion exceeds 16% involves calculating Z = (0.16 - 0.12)/√(0.12×0.88/56) ≈ 0.4, leading to a probability of about 0.3446. Similarly, for 112 items, the probability is similarly calculated and tends to decrease as sample size increases, illustrating the effect of sample size on process control decisions.

In the context of employment prospects, the sample of 21 friends, where 63% are expected to find employment initially, is modeled statistically. The expected value of successes is approximately 13.23, with a standard error of about 1.574. The sampling distribution's normal approximation allows the calculation of the probability that fewer than 53% of friends (about 11 successes) will find employment. The Z-score for this scenario is roughly -0.486, corresponding to a probability of approximately 0.313. This indicates a moderate chance that fewer than 11 friends will secure employment in the first round.

For evaluations of dealership satisfaction, with 62% of customers reporting favorably in a sample of 66, the probability of the dealership receiving between 26 and 33 favorable reports (i.e., potential fines) is calculated via normal approximation. The mean number of favorable reports is about 40.92, with a standard deviation approx 3.81. Z-scores for the bounds translate to approximately -2.3 and -1.8, rendering probabilities of about 0.0107 and 0.0359, respectively, indicating relatively low probabilities of these specific outcomes.

Finally, constructing confidence intervals for the mean 30-year fixed mortgage rate involves sampling data (mean = 5.3%, standard deviation = 0.3%) and applying the normal distribution to obtain 95% and 99% confidence intervals. These intervals, approximately (4.9%, 5.7%) and (4.7%, 6.0%), respectively, demonstrate increased uncertainty at higher confidence levels, aligning with the principle that broader intervals provide more reliable estimates.

In all these cases, the use of statistical confidence intervals, probability calculations, and hypothesis tests provides valuable insights into data, enabling informed decision-making. The key takeaway is that increasing confidence levels widens intervals and reduces the chance of Type I errors, but often at the cost of precision, highlighting the importance of choosing appropriate confidence levels based on context.

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