Question

HELP i need full diagram of 12.2 and 12.3 also please show the steps

12.1 What is measured by the estimated standard error that is used for the independent measures t statistic? 12.2 a. One sample has n = 15 with SS = 1640, and a second sample has n = 15 with SS = 1700. Find the pooled variance for the two samples. b. Compute the estimated standard error for the sample mean differences (i.e. for the two groups). If the sample mean difference (ie. Mi-M2) is 8 points, is this enough to indicate a significant difference for a two-tailed test at a..05 level? d. If the sample mean difference (ie. M1-M2) is 12 points, is this enough to indicate a significant difference for a two-tailed test at a = .05 level? c. 12.3 Siegel (1990) found that elderly people who owned dogs were less likely to pay visits to their doctors after upsetting events than were those who did not own pets. Similarly, consider the following hypothetical data. A sample of elderly dog owners is compared to a similar group (in terms of age and health) who do not own dogs. The researcher records the number of visits to the doctor during the past year for each person. The data are as follows: Control Groun

Answer 1

12.2

a)

Answer:

Pooled variance = 119.2857

Explanation:

Pooled variance

$s_p^2=\frac{SS_1+SS_2}{n_1+n_2-2}$

$s_p^2=\frac{1640+1700}{15+15-2}=119.2857$

b)

Answer:

standard error = 3.9881

Explanation:

The standard error for the difference in means

$SE=\sqrt{s_p^2\left ( \frac{1}{n_1}+\frac{1}{n_2} \right )}$

$SE=\sqrt{119.2857\left ( \frac{1}{15}+\frac{1}{15} \right )}=3.9881$

c)

Answer:

There is not sufficient evidence to conclude that the two sample means are different.

Explanation:

Test statistic

The t statistic is obtained using the following formula,

$t=\frac{M_1-M_1}{SE}=\frac{8}{3.9881}=2.0060$

P-value

The p-value for t = 2.006 is obtained from the t distribution table for the degree of freedom = n1+n2-1=28 for the two-tailed t distribution

$\text{P-value}=0.0546$

Conclusion:

Since the p-value = 0.0546 is greater than 0.05 at a 5% significance level, the null hypothesis is not rejected. hence there is not sufficient evidence to conclude that the two sample means are different.

d)

Answer:

There is sufficient evidence to conclude that the two sample means are different.

Explanation:

Test statistic

The t statistic is obtained using the following formula,

$t=\frac{M_1-M_1}{SE}=\frac{12}{3.9881}=3.0090$

P-value

The p-value for t = 3.009 is obtained from the t distribution table for the degree of freedom = n1+n2-1=28 for the two-tailed t distribution

$\text{P-value}=0.0055$

Conclusion:

Since the p-value = 0.0055 is less than 0.05 at a 5% significance level, the null hypothesis is rejected. Hence there is sufficient evidence to conclude that the two sample means are different.

12.3

a)

Answer:

There is sufficient evidence to conclude that the mean number of doctor visits between dog owners and control subjects is different.

Explanation:

The test is performed in the following steps,

Step 1: Hypothesis

The Null and Alternative Hypotheses

$H_0: \mu_1=\mu_2$

$H_0: \mu_1\neq\mu_2$

This is a two-tailed test

Step 2: Decision Rule

The t statistic is used to compare the two population means and the significance level for the test is

g= 0,05

The decision rules state the conditions that if,

,

P-value <a= 0.05, reject the null hypothesis,

Step 3: Test statistic

The t statistic is obtained using the formula,

$t=\frac{\overline{X}_1-\overline{X}_2}{\sqrt{\left ( \frac{(n_1-1)s_1^2)&plus;(n_2-1)s_2^2)}{n_1&plus;n_2-2} \right )\left ( \frac{1}{n_1}&plus;\frac{1}{n_2} \right )}}$

From the samples data values,

…	Control Group	Dog Owners
Sample mean	8.5714	5.4
Sample std dev	2.2254	1.8166
n	7	5

putting these value is above formula,

$t=2.6146$

Step 4: P-value

The P-value for the t statistic is obtained from t distribution table for the degree of freedom = n1+n2-2=7+5-2=10

$\text{P-value}=0.0258$

Step 5: Conclusion

Since the P-value = 0.0258 is less than 0.05 at a 5% significance level, the null hypothesis is rejected. Hence there is sufficient evidence to conclude that the two sample means are different.