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The following results are for independent random samples taken from two populations. $$\begin{array}{ll} \text { Sample 1 } & \text { Sample 2 } \\ n_{1}=20 & n_{2}=30 \\ \bar{x}_{1}=22.5 & \bar{x}_{2}=20.1 \\ s_{1}=2.5 & s_{2}=4.8 \end{array}$$ a. What is the point estimate of the difference between the two population means? b. What is the degrees of freedom for the \(t\) distribution? c. \(\quad\) At \(95 \%\) confidence, what is the margin of error? d. What is the \(95 \%\) confidence interval for the difference between the two population means?

Short Answer

Expert verified
a. 2.4 b. 190 c. 2.05 d. (0.35, 4.45)

Step by step solution

01

Calculate Point Estimate

The point estimate of the difference between the two population means is calculated as the difference between the sample means: \[ \bar{x}_1 - \bar{x}_2 = 22.5 - 20.1 = 2.4 \]
02

Calculate Degrees of Freedom

Since the variances of the two samples are different, we use Welch's approximation for the degrees of freedom for the \( t \) distribution in an independent samples \( t \)-test:\[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}}\]Substituting the given values: \[ s_1 = 2.5, \ s_2 = 4.8, \ n_1 = 20, \ n_2 = 30\]\[ df \approx \frac{\left(\frac{2.5^2}{20} + \frac{4.8^2}{30}\right)^2}{\frac{\left(\frac{2.5^2}{20}\right)^2}{19} + \frac{\left(\frac{4.8^2}{30}\right)^2}{29}} = \frac{\left(0.3125 + 0.768\right)^2}{\frac{0.09765625}{19} + \frac{0.0287744}{29}} \]\[ = \frac{1.0805^2}{0.005145 + 0.00099222} \approx \frac{1.167}{0.00613722} \approx 190.246\]This result is then rounded to the nearest whole number, so \(df = 190\).
03

Find Critical t-Value

With a \(95\%\) confidence level and \(df = 190\), we find the critical \(t\)-value from a \(t\)-distribution table or calculator. This value is approximately \( t^* \approx 1.972 \).
04

Calculate Margin of Error

The margin of error \(E\) is found using the formula:\[ E = t^* \times \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]Substitute the known values:\[ E = 1.972 \times \sqrt{\frac{2.5^2}{20} + \frac{4.8^2}{30}} = 1.972 \times \sqrt{0.3125 + 0.768} = 1.972 \times \sqrt{1.0805} \approx 1.972 \times 1.0395 \approx 2.05\]
05

Construct the Confidence Interval

The confidence interval for the difference between the population means is given by: \[ \left( (\bar{x}_1 - \bar{x}_2) - E, (\bar{x}_1 - \bar{x}_2) + E \right) \]\[ \left( 2.4 - 2.05, 2.4 + 2.05 \right) = \left( 0.35, 4.45 \right) \]Thus, the \(95\%\) confidence interval for the difference between the two population means is \( (0.35, 4.45) \).

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Key Concepts

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

Point Estimate
The point estimate provides a single value as an estimate of an unknown population parameter. In the context of comparing two populations, such as samples from different groups, the point estimate refers to the difference between the sample means.
This difference represents an estimate of how much one population mean deviates from the other.
In our given exercise, we calculate the point estimate by simply subtracting the mean of Sample 2 from the mean of Sample 1:
  • Sample 1 mean = 22.5
  • Sample 2 mean = 20.1
  • Point estimate = 22.5 - 20.1 = 2.4
Thus, the point estimate of the difference between the two population means is 2.4.
This number suggests that, based on our samples, the mean of the first population is 2.4 units higher than that of the second population.
Degrees of Freedom
Degrees of freedom (df) are an integral component in statistical analysis, especially when dealing with variance and mean estimates. They are essentially the number of independent values that can vary in an analysis without breaking any constraints established on the data.
In our case, while comparing two samples with possibly different variances, Welch's approximation calculates the degrees of freedom. This formula gives a more precise account when the sample variances differ.
Welch's approximation uses the formula:\[df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}}\]Here:
  • \(s_1 = 2.5\), \(s_2 = 4.8\)
  • \(n_1 = 20\), \(n_2 = 30\)
and estimates the degrees of freedom to be approximately 190 after computation, helping refine the hypothesis test results.
Welch's Approximation
Welch's approximation specifically addresses challenges when working with two independent samples having unequal variances. This method adjusts the degrees of freedom calculation in a two-sample t-test.
By not assuming equal population variances, Welch's method increases accuracy in estimating test statistics for more reliable outcomes.
In this exercise, we utilize the formula to compute degrees of freedom when sample variances are not equal, providing flexibility in statistical tests:
  • It's especially useful when sample sizes and variances differ significantly.
  • It avoids underestimating the variability and thus maintains the integrity of the test results.
The calculated degrees of freedom using Welch's approximation was approximately 190, ensuring that the confidence interval estimation is both accurate and robust.
Margin of Error
The margin of error in statistics represents the extent of potential variation in a sample estimate, offering a range that likely contains the true population parameter.
It is especially crucial in confidence intervals, as it determines the range either side of a point estimate.
To calculate the margin of error (E) for the difference between two means, we use:\[E = t^* \times \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]Here:
  • The critical t-value \(t^*\) for \(df = 190\) is approximately 1.972.
  • Substituting the variances and sample sizes, we find \(E = 2.05\).
This value indicates that the true difference in means is expected to be within 2.05 units of our point estimate, with a 95% confidence.

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Most popular questions from this chapter

Four different paints are advertised as having the same drying time. To check the manufacturer's claims, five samples were tested for each of the paints. The time in minutes until the paint was dry enough for a second coat to be applied was recorded. The following data were obtained. $$\begin{array}{cccc} \text { Paint 1 } & \text { Paint 2 } & \text { Paint 3 } & \text { Paint 4 } \\\ 128 & 144 & 133 & 150 \\ 137 & 133 & 143 & 142 \\ 135 & 142 & 137 & 135 \\ 124 & 146 & 136 & 140 \\ 141 & 130 & 131 & 153 \end{array}$$ At the \(\alpha=.05\) level of significance, test to see whether the mean drying time is the same for each type of paint.

Three different methods for assembling a product were proposed by an industrial engineer. To investigate the number of units assembled correctly with each method, 30 employees were randomly selected and randomly assigned to the three proposed methods in such a way that each method was used by 10 workers. The number of units assembled correctly was recorded, and the analysis of variance procedure was applied to the resulting data set. The following results were obtained: \(\mathrm{SST}=10,800 ; \mathrm{SSTR}=4560\) a. Set up the ANOVA table for this problem. b. Use \(\alpha=.05\) to test for any significant difference in the means for the three assembly methods.

Consider the following hypothesis test. $$\begin{array}{l} H_{0}: \mu_{1}-\mu_{2}=0 \\ H_{\mathrm{a}}: \mu_{1}-\mu_{2} \neq 0 \end{array}$$ The following results are for two independent samples taken from the two populations. $$\begin{array}{ll} \text { Sample 1 } & \text { Sample 2 } \\ n_{1}=80 & n_{2}=70 \\ \bar{x}_{1}=104 & \bar{x}_{2}=106 \\ \sigma_{1}=8.4 & \sigma_{2}=7.6 \end{array}$$ a. What is the value of the test statistic? b. What is the \(p\) -value? c. With \(\alpha=.05,\) what is your hypothesis testing conclusion?

The National Association of Home Builders provided data on the cost of the most popular home remodeling projects. Sample data on cost in thousands of dollars for two types of remodeling projects are as follows. $$\begin{array}{cccc} \text { Kitchen } & \text { Master Bedroom } & \text { Kitchen } & \text { Master Bedroom } \\ 25.2 & 18.0 & 23.0 & 17.8 \\ 17.4 & 22.9 & 19.7 & 24.6 \\ 22.8 & 26.4 & 16.9 & 21.0 \\ 21.9 & 24.8 & 21.8 & \\ 19.7 & 26.9 & 23.6 & \end{array}$$ a. Develop a point estimate of the difference between the population mean remodeling costs for the two types of projects. b. Develop a \(90 \%\) confidence interval for the difference between the two population means.

Three different assembly methods have been proposed for a new product. A completely randomized experimental design was chosen to determine which assembly method results in the greatest number of parts produced per hour, and 30 workers were randomly selected and assigned to use one of the proposed methods. The number of units produced by each worker follows. $$\begin{array}{rrr} & \text { Method } & \\ \mathbf{A} & \mathbf{B} & \mathbf{C} \\ 97 & 93 & 99 \\ 73 & 100 & 94 \\ 93 & 93 & 87 \\ 100 & 55 & 66 \\ 73 & 77 & 59 \\ 91 & 91 & 75 \\ 100 & 85 & 84 \\ 86 & 73 & 72 \\ 92 & 90 & 88 \\ 95 & 83 & 86 \end{array}$$ Use these data and test to see whether the mean number of parts produced is the same with each method. Use \(\alpha=.05\)

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