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Two levels (low and high) of insulin doses are given to two groups of diabetic rats to check the insulinbinding capacity, yielding the following data: $$\begin{array}{llll}\text { Low dose: } & n_{1}=8 & x_{1}-1.98 & s_{1}=0.51 \\\ \text { High dose: } & n_{2}=13 & \bar{x}_{2}=1.30 & 8 \mathrm{a}=0.35\end{array}$$ Assume that both variances are equal. Give a \(95 \%\) confidence interval for the difference of the true average insulin-binding capacity between two samples.

Short Answer

Expert verified
The 95% confidence interval for the difference of the true average insulin-binding capacity between the two samples is the calculated interval from step 4.

Step by step solution

01

Calculate pooled variance

Pooled variance is calculated using the following formula: \[s_p^2 = \frac{(n_{1}-1)s_{1}^2+(n_{2}-1)s_{2}^2}{n_{1}+n_{2}-2}\] Substituting for \(n_{1} = 8\), \(s_{1} = 0.51\), \(n_{2} = 13\) and \(s_{2} = 0.35\), we find \(s_p^2\).
02

Calculate standard error of difference

Calculate the standard error of the difference using the equation: \[SE = s_p\sqrt{1/n_{1} + 1/n_{2}}\]
03

Obtain t-critical

For a 95% confidence interval, and degrees of freedom (df) equal to \(n_{1}+n_{2}-2 = 8+13-2=19\), we find from t-distribution table that t-critical (t*) is approximately 2.093.
04

Calculate confidence interval

Now, we calculate the confidence interval using the formula: \[(\bar{x}_{1} - \bar{x}_{2}) \pm t^*SE\] Substitute in the given values: \(\bar{x}_{1} = 1.98\), \(\bar{x}_{2} = 1.30\) and the calculated values for t* and SE. This will give the 95% confidence interval.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pooled Variance
When comparing two groups, such as different dosages of insulin on rats, we often assume that the variability within each group is the same. This common variability is known as 'pooled variance'. To calculate pooled variance, we combine the variances of both groups, weighing them by the number of observations in each group minus one, which represents their degrees of freedom.

The formula to calculate pooled variance, denoted as \( s_p^2 \), is:\[ s_p^2 = \frac{(n_{1}-1)s_{1}^2 + (n_{2}-1)s_{2}^2}{n_{1} + n_{2} - 2} \]Here, \( n_{1} \) and \( n_{2} \) represent the sample sizes of each group, while \( s_{1} \) and \( s_{2} \) are their respective sample standard deviations. In our insulin dose example, this calculation helps us to derive a standard measure of variability that we'll use in further computations, like estimating the standard error.
Standard Error
The standard error (SE) measures the accuracy with which a sample represents a population. In our case, it's about how well our sample means estimate the true mean difference in insulin-binding capacity. After pooling the variances with the earlier formula, we may calculate the standard error of the difference between two means using this formula:\[ SE = s_p\sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}} \]

This standard error helps quantify the uncertainty or 'error' in our estimate due to the simple fact that we're working with samples, not entire populations. It essentially tells us that if we were to conduct the same experiment multiple times, we'd expect the difference in means to vary by about this standard error amount. Thus, it's an essential part of forming a confidence interval as it directly influences how 'wide' or 'narrow' that interval will be.
T-distribution
The t-distribution is particularly useful when dealing with small sample sizes or when the population variance is unknown. It resembles the normal distribution but has fatter tails, allowing for more variability that's typical in small samples. As our sample size grows, the t-distribution approaches the normal distribution.

In the context of confidence intervals, the t-distribution is used to find the t-critical value, or \( t^* \), which corresponds to the desired confidence level—in our rat study, it's 95%. The t-critical value is a factor that, when multiplied by the standard error, gives us the range around the sample mean difference that we can be 'confident' contains the true population mean difference. Finding the correct t-critical value requires knowing the degrees of freedom, which we'll discuss next.
Degrees of Freedom
Degrees of freedom (df) can be thought of as the number of independent pieces of information remaining after estimating certain parameters. In the case of variance, degrees of freedom are associated with the number of values that are free to vary. When calculating our pooled variance, we use \( n_{1} + n_{2} - 2 \) as our degrees of freedom, which represents the total number of observations minus the number of group means estimated.

The concept of degrees of freedom is crucial when referencing tables for the t-distribution to locate the appropriate t-critical value. It's directly related to the shape of the t-distribution; the more degrees of freedom there are, the closer the distribution resembles the standard normal distribution. In our exercise, the degrees of freedom influence the 'confidence' we have in our interval estimate by helping to determine the multiplicative factor (t-critical value) we use to adjust our margin of error.

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Most popular questions from this chapter

A specific labor union is becoming defensive about gross absenteeism by its members. The union decided to check this by monitoring a random sample of its members. The union leaders had always claimed that in a typical month, \(95 \%\) of its members are absent less than 10 hours per month. Use the data to respond to this claim. Use a one-sided tolerance limit and choose the confidence level to be \(99 \%\). Be sure to interpret what you learned from the tolerance limit calculation. The number of members in this sample was 300 . The number of hours absent was recorded for each of the 300 members. The results were \(x=6.5\) hours and \(s=2.5\) hours.

(a) According to a report in the Roanoke Times \& World-News, approximately \(2 / 3\) of the 1600 adults polled by telephone said they think the space shuttle program is a good investment for the country. Find a \(95 \%\) confidence interval for the proportion of American adults who think the space shuttle program is a good investment for the country. (b) What can we assert with \(95 \%\) confidence about the possible size of our error if we estimate the proportion of American adults who think the space shuttle program is a good investment to be \(2 / 3 ?\)

Ten engineering schools in the United States were surveyed. The sample contained 250 electrical engineers, 80 being women; 175 chemical engineers, 40 being women. Compute a \(90 \%\) confidence interval for the difference between the proportion of women in these two fields of engineering. Is there a significant difference between the two proportions?

An anthropologist is interested in the proportion of individuals in two Indian tribes with double occipital hair whorls. Suppose that independent samples are taken from each of the two tribes, and it is found that 24 of 100 Indians from tribe \(A\) and 36 of 120 Indians from tribe \(B\) possess this characteristic. Construct a \(95 \%\) confidence interval for the difference \(p_{B}-p_{A}\) between the proportions of these two tribes with occipital hair whorls.

A machine is used to fill boxes of product in an assembly line operation. Much concern centers around the variability in the number of ounces of product in the box. The standard deviation in weight of product is known to be 0.3 ounces. An improvement is implemented after which a random sample of 20 boxes are selected and the sample variance is found to be 0.045 ounces. Find a \(95 \%\) confidence interval on the variance in the weight of the product. Does it appear from the range of the confidence interval that the improvement of the process enhanced quality as far as variability is concerned? Assume normality on the distribution of weight of product.

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