Question

Use the t-distribution to find a confidence interval for a difference in means μ1-μ2 given the relevant sample results. Give the best estimate for μ1-μ2, the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed.

A 95% confidence interval for μ1-μ2 using the sample results x¯1=79.0, s1=10.5, n1=35 and x¯2=65.8, s2=7.2, n2=20. Enter the exact answer for the best estimate and round your answers for the margin of error and the confidence interval to two decimal places.

a. Best estimate = _______
b. Margin of error = _________
c. Confidence interval ________

Answer 1

Answer:

(a) The best estimate of $\mu_{1}-\mu_{2}$ is 13.2.

(b) The margin of error is 5.30.

(c) The 95% confidence interval for the difference between two means is (7.90, 18.50).

Step-by-step explanation:

The (1 - α)% confidence interval for the difference between population means using a t-interval is:

$CI=(\bar x_{1}-\bar x_{2})\pm t_{\alpha/2, (n_{1}+n_{2}-2)} \sqrt{S_{p}^{2}[\frac{1}{n_{1}}+\frac{1}{n_{2}}]}}$

(a)

A point estimate of a parameter (population) is a distinct value used for the estimation the parameter (population). For instance, the sample mean $\bar x$ is a point-estimate of the population mean μ.

Similarly the point estimate of the difference between two means is:

$\bar x_{1}-\bar x_{2}$

Compute the point estimate of $\mu_{1}-\mu_{2}$ as follows:

$E(\mu_{1}-\mu_{2})=\bar x_{1}-\bar x_{2}\\=79.0-65.8\\=13.2$

Thus, the best estimate of $\mu_{1}-\mu_{2}$ is 13.2.

(b)

Compute the pooled variance as follows:

$S_{p}^{2}=\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}=\frac{(35-1)10.5^{2}+(20-1)7.2^{2}}{35+20-2}=89.311$

Compute the critical value of t as follows:

$t_{\alpha/2, (n_{1}+n_{2}-2)}=t_{0.05/2, (35+20-2)}=t_{0.025, 53}=2.00$

*Use a t-table.

Compute the margin of error as follows:

$MOE=t_{\alpha/2, (n_{1}+n_{2}-2)} \sqrt{S_{p}^{2}[\frac{1}{n_{1}}+\frac{1}{n_{2}}]}}$

          $=2.00\times\sqrt{89.311\times [\frac{1}{35}+\frac{1}{20}]} \\=5.298\\\approx5.30$

Thus, the margin of error is 5.30.

(c)

Compute the 95% confidence interval for the difference between two means as follows:

$CI=(\bar x_{1}-\bar x_{2})\pm MOE$

      $=13.2\pm 5.298\\=(7.902, 18.498)\\\approx (7.90, 18.50)$

Thus, the 95% confidence interval for the difference between two means is (7.90, 18.50).

Answer 2

Using the t-distribution, it is found that:

a) The best estimate is of 13.8.

b) The margin of error is of 4.81.

c) The confidence interval is (8.99, 18.61).

The standard errors are given by:

$s_1 = \frac{10.5}{\sqrt{35}} = 1.7748$

$s_2 = \frac{7.2}{\sqrt{20}} = 1.61$

Item a:

The distribution of the difference is given by:

$\overline{x} = \mu_1 - \mu_2 = 79 - 65.8 = 13.2$

$s = \sqrt{s_1^2 + s_2^2} = \sqrt{1.7748^2 + 1.61^2} = 2.39625$

Hence: The best estimate is of 13.8.

Item b:

The margin of error is:

$M = ts$

t is the critical value, which in this problem is for a two-tailed 95% confidence interval, with 35 + 20 - 2 = 53 df, hence t = 2.0057.

Then:

$M = ts = 2.0057(2.39625) = 4.81$

The margin of error is of 4.81.

Item c:

The confidence interval is the best estimate plus/minus the margin of error, hence:

$\overline{x} - M = 13.8 - 4.81 = 8.99$

$\overline{x} + M = 13.8 + 4.81 = 18.61$

The confidence interval is (8.99, 18.61).

A similar problem is given at https://brainly.com/question/16184804