Answer :
Sure, let's walk through each part of the question step-by-step using the provided data and results.
### a. Compute the Sample Means
First, let's compute the means for the two samples.
Sample 1: 10, 7, 13, 7, 9, 8
To find the mean:
1. Sum the values: [tex]\(10 + 7 + 13 + 7 + 9 + 8 = 54\)[/tex]
2. Divide by the number of values: [tex]\( \frac{54}{6} ≈ 9 \)[/tex]
Therefore, the mean of Sample 1 ([tex]\(x_1\)[/tex]) is [tex]\(9\)[/tex].
Sample 2: 8, 7, 8, 4, 6, 9
To find the mean:
1. Sum the values: [tex]\(8 + 7 + 8 + 4 + 6 + 9 = 42\)[/tex]
2. Divide by the number of values: [tex]\( \frac{42}{6} ≈ 7 \)[/tex]
Therefore, the mean of Sample 2 ([tex]\(x_2\)[/tex]) is [tex]\(7\)[/tex].
So,
[tex]\[ \begin{array}{l} x_1 = 9 \\ x_2 = 7 \end{array} \][/tex]
### b. Compute the Sample Standard Deviations
Next, let's compute the standard deviations for the two samples, rounded to two decimal places.
Sample 1: 10, 7, 13, 7, 9, 8
The sample standard deviation ([tex]\(s_1\)[/tex]) is approximately [tex]\(2.28\)[/tex].
Sample 2: 8, 7, 8, 4, 6, 9
The sample standard deviation ([tex]\(s_2\)[/tex]) is approximately [tex]\(1.79\)[/tex].
So,
[tex]\[ \begin{array}{l} s_1 = 2.28 \\ s_2 = 1.79 \end{array} \][/tex]
### c. Point Estimate of the Difference Between the Two Population Means
The point estimate of the difference between the two population means is given by the difference between the two sample means:
[tex]\[ \text{Point Estimate} = x_1 - x_2 = 9 - 7 = 2 \][/tex]
### d. 90% Confidence Interval Estimate of the Difference Between the Two Population Means
The 90% confidence interval estimate for the difference between the two population means can be found using the following:
1. Calculate the standard error of the difference:
[tex]\[ SE = \sqrt{ \left( \frac{s_1^2}{n_1} \right) + \left( \frac{s_2^2}{n_2} \right) } = \sqrt{ \left( \frac{2.28^2}{6} \right) + \left( \frac{1.79^2}{6} \right) } \][/tex]
2. Determine the degrees of freedom:
[tex]\[ df = n_1 + n_2 - 2 = 6 + 6 - 2 = 10 \][/tex]
3. Find the critical t-value for a 90% confidence level with 10 degrees of freedom. This value is [tex]\( t_{critical} = t_{0.95, 10} \)[/tex].
4. Calculate the margin of error:
[tex]\[ ME = t_{critical} \times SE \][/tex]
5. Construct the confidence interval:
[tex]\[ (\text{Point Estimate} - ME, \text{Point Estimate} + ME) \][/tex]
From the results, the confidence interval is approximately from [tex]\(-0.14\)[/tex] to [tex]\(4.14\)[/tex].
So, the confidence interval is:
[tex]\[ \begin{array}{ll} -0.14 & \\ 4.14 & \end{array} \][/tex]
Putting it all together:
#### a. Compute the two sample means. (to nearest whole number)
[tex]\[ x_1 = 9, x_2 = 7 \][/tex]
#### b. Compute the two sample standard deviations. (to 2 decimals)
[tex]\[ s_1 = 2.28, s_2 = 1.79 \][/tex]
#### c. What is the point estimate of the difference between the two population means?
[tex]\[ 2 \][/tex]
#### d. What is the [tex]\(90\%\)[/tex] confidence interval estimate of the difference between the two population means? (to 2 decimals and enter negative value as negative number)
[tex]\[ -0.14, 4.14 \][/tex]
### a. Compute the Sample Means
First, let's compute the means for the two samples.
Sample 1: 10, 7, 13, 7, 9, 8
To find the mean:
1. Sum the values: [tex]\(10 + 7 + 13 + 7 + 9 + 8 = 54\)[/tex]
2. Divide by the number of values: [tex]\( \frac{54}{6} ≈ 9 \)[/tex]
Therefore, the mean of Sample 1 ([tex]\(x_1\)[/tex]) is [tex]\(9\)[/tex].
Sample 2: 8, 7, 8, 4, 6, 9
To find the mean:
1. Sum the values: [tex]\(8 + 7 + 8 + 4 + 6 + 9 = 42\)[/tex]
2. Divide by the number of values: [tex]\( \frac{42}{6} ≈ 7 \)[/tex]
Therefore, the mean of Sample 2 ([tex]\(x_2\)[/tex]) is [tex]\(7\)[/tex].
So,
[tex]\[ \begin{array}{l} x_1 = 9 \\ x_2 = 7 \end{array} \][/tex]
### b. Compute the Sample Standard Deviations
Next, let's compute the standard deviations for the two samples, rounded to two decimal places.
Sample 1: 10, 7, 13, 7, 9, 8
The sample standard deviation ([tex]\(s_1\)[/tex]) is approximately [tex]\(2.28\)[/tex].
Sample 2: 8, 7, 8, 4, 6, 9
The sample standard deviation ([tex]\(s_2\)[/tex]) is approximately [tex]\(1.79\)[/tex].
So,
[tex]\[ \begin{array}{l} s_1 = 2.28 \\ s_2 = 1.79 \end{array} \][/tex]
### c. Point Estimate of the Difference Between the Two Population Means
The point estimate of the difference between the two population means is given by the difference between the two sample means:
[tex]\[ \text{Point Estimate} = x_1 - x_2 = 9 - 7 = 2 \][/tex]
### d. 90% Confidence Interval Estimate of the Difference Between the Two Population Means
The 90% confidence interval estimate for the difference between the two population means can be found using the following:
1. Calculate the standard error of the difference:
[tex]\[ SE = \sqrt{ \left( \frac{s_1^2}{n_1} \right) + \left( \frac{s_2^2}{n_2} \right) } = \sqrt{ \left( \frac{2.28^2}{6} \right) + \left( \frac{1.79^2}{6} \right) } \][/tex]
2. Determine the degrees of freedom:
[tex]\[ df = n_1 + n_2 - 2 = 6 + 6 - 2 = 10 \][/tex]
3. Find the critical t-value for a 90% confidence level with 10 degrees of freedom. This value is [tex]\( t_{critical} = t_{0.95, 10} \)[/tex].
4. Calculate the margin of error:
[tex]\[ ME = t_{critical} \times SE \][/tex]
5. Construct the confidence interval:
[tex]\[ (\text{Point Estimate} - ME, \text{Point Estimate} + ME) \][/tex]
From the results, the confidence interval is approximately from [tex]\(-0.14\)[/tex] to [tex]\(4.14\)[/tex].
So, the confidence interval is:
[tex]\[ \begin{array}{ll} -0.14 & \\ 4.14 & \end{array} \][/tex]
Putting it all together:
#### a. Compute the two sample means. (to nearest whole number)
[tex]\[ x_1 = 9, x_2 = 7 \][/tex]
#### b. Compute the two sample standard deviations. (to 2 decimals)
[tex]\[ s_1 = 2.28, s_2 = 1.79 \][/tex]
#### c. What is the point estimate of the difference between the two population means?
[tex]\[ 2 \][/tex]
#### d. What is the [tex]\(90\%\)[/tex] confidence interval estimate of the difference between the two population means? (to 2 decimals and enter negative value as negative number)
[tex]\[ -0.14, 4.14 \][/tex]
