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When dealing with statistical data, the t-distribution is essential when the sample size is small (usually less than 30) and the population standard deviation is unknown. Unlike the normal distribution, the t-distribution accounts for the added variability that small samples introduce to a dataset. This is crucial when estimating population parameters, like a mean, from a single sample set.

Using a t-distribution instead of a normal distribution gives a more accurate confidence interval. The t-distribution has thicker tails, which means it is more likely to produce values far from the mean compared to a normal distribution. This characteristic compensates for the increased uncertainty inherent in smaller samples.

In our vitamin C measurement problem, the sample size is 8, which is quite small. Thus, it is appropriate to use the t-distribution to find the critical t-value necessary for constructing the confidence interval for the mean vitamin C content.

The sample mean, denoted by \( \bar{x} \), is a simple yet effective statistic used to estimate the average of a population. In essence, it gives us a central value of the data set collected from a particular sample.

Calculating the sample mean involves adding up all the observed values in the sample and dividing the total by the number of observations. In the case of the vitamin C measurements, the eight sample observations add up to 180 mg/100 g. By dividing this sum by the number of samples, which is 8, we find the sample mean to be 22.5 mg/100 g.

This calculated mean is crucial because it acts as an estimate of the population mean, which helps us construct the confidence interval to understand variability and the assurance we have around this mean.

To understand how dispersed the data points are around the sample mean, we compute the sample standard deviation, denoted by \( s \). This statistic provides insight into the variability of the sample in comparison to the mean.

To find the sample standard deviation, we first calculate each data point's squared deviation from the sample mean, add these squared deviations, and divide by \( n-1 \) since we are working with a sample rather than a complete population. Finally, taking the square root of this value gives us the standard deviation.

In the context of the vitamin C data, the squared deviations added up to 360, and dividing by 7 gives a variance of approximately 51.43 mg/100 g. The square root of this variance yields a sample standard deviation of approximately 7.17 mg/100 g. The sample standard deviation is key for calculating the margin of error and, consequently, the confidence interval.

Measuring vitamin C content in food samples is an important assessment of nutritional value. In the provided problem, the focus is on the vitamin C levels in corn-soy blend (CSB), a fortified food used in relief efforts.

A random sample of 8 measurements represents the vitamin C content, and this sample provides data to analyze the mean vitamin C levels. By constructing a 95\(\%\) confidence interval, we identify the range likely to encompass the true population mean of vitamin C content in the CSB.

We found the interval to be between 16.51 and 28.49 mg/100 g based on our sample data. This finding offers insight into the adequacy of vitamin C levels, helping to inform whether these values meet the nutritional standards desired for the relief efforts.