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The following information was obtained from two independent samples selected from two normally distributed populations with unknown but equal standard deviations. $$ \begin{array}{lll} n_{1}=21 & \bar{x}_{1}=13.97 & s_{1}=3.78 \\ n_{2}=20 & \bar{x}_{2}=15.55 & s_{2}=3.26 \end{array} $$ a. What is the point estimate of \(\mu_{1}-\mu_{2} ? \quad\) b. Construct a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\).

Short Answer

Expert verified
a. The point estimate of \(\mu_{1}-\mu_{2}\) is -1.58. b. The 95% confidence interval for \(\mu_{1}-\mu_{2}\) is (-3.73, 0.57).

Step by step solution

01

Calculate the Point Estimate

The point estimate of difference of means \(\mu_{1}-\mu_{2}\) is simply the difference of sample means, i.e., \(\bar{x}_{1} - \(\bar{x}_{2}. Substituting the values we get 13.97 - 15.55 = -1.58.
02

Calculate the Pooled Standard Deviation

Next, we calculate the pooled standard deviation using the formula \(\sqrt{\((s_{1}^2/n1) + (s_{2}^2/n2)\)}. Substituting the given values, we get \(\sqrt{\((3.78^2/21) + (3.26^2/20)\} = 1.075.
03

Find the t value for 95% confidence level

Look up the t-distribution table with \(df= n_{1}+n_{2}-2 = 21+20-2 = 39\) degrees of freedom for 95% confidence interval (two-tailed test), we get \(t=2.022\)
04

Construct the Confidence Interval

The formula for a confidence interval is (point estimate - margin of error, point estimate + margin of error). The margin of error can be calculated by multiplying the standard deviation by the t-value from step 3. This yields -1.58-2.022*1.075, -1.58+2.022*1.075, which simplifies to (-3.73, 0.57).

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Key Concepts

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

Point Estimate
In statistics, a point estimate refers to a single value that serves as an estimate of an unknown population parameter. When estimating the difference between two population means, the point estimate is found by calculating the difference between the sample means. This specific calculation gives us a close approximation of \(\mu_{1} - \mu_{2}\), the true average difference between the populations.

For instance, in our exercise, we have two sample means: \(\bar{x}_{1} = 13.97\) and \(\bar{x}_{2} = 15.55\). The point estimate of \(\mu_{1} - \mu_{2}\) is simply the subtraction of these two numbers: \(-1.58\). This point estimate is critical in constructing confidence intervals, as it centers around which range of possible values the true population parameter could lie.
Pooled Standard Deviation
The pooled standard deviation is a technique used in the context of statistical analysis to combine the variances of two independent samples. When population standard deviations are unknown yet assumed to be equal, this method is beneficial in comparing two group means.

Calculated with the following formula:\[s_{p} = \sqrt{\frac{(n_{1} - 1)s_{1}^{2} + (n_{2} - 1)s_{2}^{2}}{n_{1} + n_{2} - 2}}\] This formula accounts for the individual variances while weighing them by their sample sizes. The pooled standard deviation \(s_{p} \) provides the variability estimate associated with both groups, which is then used in further calculations, such as determining the confidence interval. For our samples, substituting the values gives us a pooled standard deviation of \(\approx 1.075\).
T-Distribution
The t-distribution is a type of probability distribution that is often used in statistical analysis, especially when dealing with small sample sizes. Unlike the normal distribution, it has thicker tails, which means it can account for the increased variability inherent in small samples.

In our scenario, we look up the t-distribution table to find the critical t-value for a 95% confidence interval, given the degrees of freedom. This t-value is crucial as it adjusts our margin of error for the sample size and variability estimates, ensuring our confidence interval is accurate. With 39 degrees of freedom in our exercise, the required t-value is approximately \(2.022\). This is critical in multiplying with the pooled standard deviation to find the margin of error.
Degree of Freedom
Degrees of freedom (df) refer to the number of independent values or quantities that can be assigned to a statistical distribution. It represents the number of values in the final calculation of a statistic that are free to vary.

In the construction of a confidence interval, calculating the degrees of freedom allows us to determine the appropriate critical value from a t-distribution. For two independent samples, the df is often calculated as: \(n_{1} + n_{2} - 2\), which takes into account the sample sizes involved. In our example, we compute the degrees of freedom as \(21 + 20 - 2 = 39\).

A correct understanding of degrees of freedom ensures that statistical inference is precise, as it directly impacts the width of the confidence interval through its effect on the critical t-value.

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Most popular questions from this chapter

In a random sample of 800 men aged 25 to 35 years, \(24 \%\) said they live with one or both parents. In another sample of 850 women of the same age group, \(18 \%\) said that they live with one or both parents. a. Construct a \(95 \%\) confidence interval for the difference between the proportions of all men and all women aged 25 to 35 years who live with one or both parents. b. Test at the \(2 \%\) significance level whether the two population proportions are different. c. Repeat the test of part b using the \(p\) -value approach.

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An economist was interested in studying the impact of the recession on dining out, including drive-thru meals at fast food restaurants. A random sample of forty-eight families of four with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week indicated that they reduced their spending on dining out by an average of \(\$ 31.47\) per week, with a sample standard deviation of \(\$ 10.95\). Another random sample of 42 families of five with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week reduced their spending on dining out by an average \(\$ 35.28\) per week, with a sample standard deviation of \(\$ 12.37\). (Note that the two groups of families are differentiated by the number of family members.) Assume that the distributions of reductions in weekly dining-out spendings for the two groups have the same population standard deviation. a. Construct a \(90 \%\) confidence interval for the difference in the mean weekly reduction in diningout spending levels for the two populations. b. Using the \(5 \%\) significance level, can you conclude that the average weekly spending reduction for all families of four with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week is less than the average weekly spending reduction for all families of five with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week?

A new type of sleeping pill is tested against an older, standard pill. Two thousand insomniacs are randomly divided into two equal groups. The first group is given the old pill, and the second group receives the new pill. The time required to fall asleep after the pill is administered is recorded for each person. The results of the experiment are given in the following table, where \(\bar{x}\) and \(s\) represent the mean and standard deviation, respectively, for the times required to fall asleep for people in each group after the pill is taken. $$ \begin{array}{lcc} \hline & \begin{array}{c} \text { Group 1 } \\ \text { (Old PilI) } \end{array} & \begin{array}{c} \text { Group 2 } \\ \text { (New Pill) } \end{array} \\ \hline n & 1000 & 1000 \\ \bar{x} & 15.4 \text { minutes } & 15.0 \text { minutes } \\ s & 3.5 \text { minutes } & 3.0 \text { minutes } \\ \hline \end{array} $$ Consider the test of hypothesis \(H_{0}: \mu_{1}-\mu_{2}=0\) versus \(H_{1}: \mu_{1}-\mu_{2}>0\), where \(\mu_{1}\) and \(\mu_{2}\) are the mean times required for all potential users to fall asleep using the old pill and the new pill, respectively. a. Find the \(p\) -value for this test. b. Does your answer to part a indicate that the result is statistically significant? Use \(\alpha=.025\). c. Find the \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). d. Does your answer to part c imply that this result is of great practical significance?

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