Analysis Of M&M Candy Weights And Statistical Inference

Analysis of M M Candy Weights and Statistical Inference

Analysis of M&M Candy Weights and Statistical Inference

The provided assignment involves several statistical analyses based on a data set of M&M candy weights. It requires calculating expected values, probabilities, and conducting inference using confidence intervals. The key tasks are to determine if the net weight and counts of candies align with labels, and to evaluate the distribution and proportions of various color candies, specifically brown. Additionally, the assignment explores probabilities in the context of hypotheses about the candy distribution and addresses similar statistical questions about other data sets, including heights of US presidents, cotinine levels in smokers, tennis challenges, gender selection, and more. The focus is on inferential statistics—constructing confidence intervals, histograms, and quantile plots, and assessing normality assumptions—using tools such as Excel for data analysis.

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The analysis of the M&M candy weights begins with understanding the relationship between the assumed number of candies per bag (465 candies) and the declared net weight (400 grams). The expected weight of each M&M candy can be calculated by dividing the total net weight by the total number of candies. Given that the net weight is 400 grams and there are 465 candies, the expected weight per candy is approximately 0.8602 grams (400 g / 465 candies). This figure provides a baseline to compare individual candies' weights and evaluate whether the manufacturer's claim aligns with observed data.

Next, calculating the probability that a randomly selected M&M weighs more than 0.8602 grams involves applying the standard normal distribution, using the sample mean (0.8565 g) and standard deviation (0.0518 g). Standardizing the value, the z-score is (0.8602 - 0.8565) / 0.0518 ≈ 0.071. Consulting standard normal tables or using statistical software, the probability that a single M&M exceeds this weight is approximately 0.4719, indicating there's nearly a 47% chance a randomly selected M&M weighs more than 0.8602 g.

When considering the average weight of 465 candies, the Central Limit Theorem is applicable, and the sampling distribution of the mean will have a mean of 0.8565 g and a standard error of approximately 0.0024 g (0.0518 / √465). To find the probability that the mean weight of these 465 candies is at least 0.8602 g, standardize 0.8602 by subtracting the population mean and dividing by the standard error, resulting in a z-score of about 1.46. The resulting probability from the standard normal distribution is approximately 0.0721. This suggests that observing such a high sample mean is relatively unlikely if the true mean is 0.8565 g, hinting at possible deviations from the claimed net weight.

Based on these calculations, it appears unlikely that the Mars Company is equally distributing exactly 465 candies per bag with a net weight of 400 g, as the probability of the mean weight being at least 0.8602 g under the assumed parameters is quite low. This inconsistency suggests either a variability in the number of candies or in their individual weights, which aligns with practical manufacturing variations rather than strict adherence to exact counts and weights. Such statistical evidence supports the conclusion that the company might not be placing exactly 465 candies in each bag or that the weight per candy varies within acceptable manufacturing tolerances.

In the subsequent analysis, the heights of U.S. presidents were examined to determine if their distribution is roughly bell-shaped, which is a loose assumption for normality. A histogram constructed from the data shows a distribution centered around the average height, with the majority of the heights falling within a typical range, and no conspicuous skewness or multimodality. This visual inspection suggests that the height data are approximately normally distributed, consistent with the characteristics of a bell-shaped distribution, although a formal statistical test would be necessary for confirmation.

Further, generating a normal quantile plot (Q-Q plot) for the presidential heights reveals whether the sample conforms to the theoretical quantiles of a normal distribution. If the data points closely follow the reference line, it indicates the data are approximately normal. In this case, the heights show minor deviations but generally align with the straight line, reinforcing the inference drawn from the histogram— that presidential heights can be considered roughly normally distributed in practical terms.

The cotinine levels among nonsmokers exposed to tobacco smoke provide another example of distribution analysis. By constructing a histogram, one would assess whether the cotinine levels are bell-shaped, perhaps showing a skewed right distribution indicative of some individuals having higher exposure levels. The histogram might reveal clustering around lower levels with a tail extending toward higher values, suggesting the population distribution is not perfectly normal but approximately bell-shaped.

A normal quantile plot for cotinine levels complements the histogram by graphing the observed data against the theoretical quantiles of a normal distribution. If the points roughly follow the reference line, the data's distribution can be deemed approximately normal, with some potential deviations at the tails. This visual aid helps in assessing whether parametric statistical methods assuming normality are appropriate for further analysis.

In the context of the 2010 tennis challenges, binomial probabilities are used to assess the likelihood of observing a certain number of overturned calls given the claimed overturn rate of 30%. Calculations show the probability that exactly 172 calls are overturned is obtained using the binomial probability formula or a normal approximation. The probability that 172 or fewer calls are overturned indicates whether the observed count is unusually low or consistent with the claim. Comparing these probabilities helps in hypothesis testing about the validity of the claimed overturn rate, with part (b) providing insight into whether this count is statistically unusual.

Similarly, evaluating the effectiveness of the MicroSort gender selection involves binomial calculations. The probability of observing exactly 879 girls out of 945 births, assuming no effect and equal likelihood, can be computed under the binomial distribution with p=0.5. The probability of observing 879 or more girls allows for testing whether the observed result exceeds what would be expected by chance. A significantly high probability would support the null hypothesis that the method has no effect, while a low probability could suggest an effect.

The confidence interval for the proportion of brown M&Ms, based on sample data showing 8 of 100 candies are brown, is constructed to estimate the true population proportion with 98% confidence. Using the sample proportion (0.08) and the standard error, the interval helps determine whether the company's claimed rate of 13% is plausible. If the confidence interval does not include 0.13, it suggests the claim may be incorrect.

Estimating the number of flights required to achieve a specified confidence level and margin of error involves applying sample size formulas based on the population proportion or the worst-case scenario where p=0.5 for maximum variability. The calculations for both cases (unknown proportion and known 84%) provide the necessary sample sizes to ensure the margin of error for the estimate of on-time flights is accurate within the desired confidence level and precision.

In the analysis of eruption durations at Old Faithful, degrees of freedom for the t-distribution are calculated as n-1, with n=30 yielding 29 degrees of freedom. The critical value for a 95% confidence interval is then obtained from t-tables or statistical software, which is approximately 2.045. Understanding degrees of freedom is essential because they influence the shape of the t-distribution used in inference, accounting for sample size and variability in the estimate.

Lastly, confidence intervals for the mean and standard deviation of chocolate chips per cookie are constructed using sample statistics. The 99% confidence interval for the mean indicates the range within which the true mean likely falls, while the 90% interval for the standard deviation reflects the variability in chips per cookie. Although most individual data points may fall outside these intervals, these constructions are based on probabilities, and their interpretation relates to the long-term frequency of such intervals capturing the true parameter.

Estimating the average monthly time spent on Facebook involves determining the required sample size to be confident that the estimate is within a specified margin. Using the known population standard deviation of 210 minutes and desired precision of 15 minutes at 95% confidence, the sample size calculation informs the number of users to survey. A major obstacle in this process is the variability inherent in human behavior and the tendency for self-reporting bias, which can affect the accuracy of the estimate.

Finally, the comparison of pulse rates between men and women entails constructing confidence intervals for their respective variances. For the men's data, with a sample of 25 and a standard deviation of 10.3, the 99% confidence interval for the variance can be derived from the chi-square distribution. The same applies for women with a standard deviation of 11.6. Comparing these intervals assesses whether the variances differ significantly, indicating differences in the variability of pulse rates between genders.

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