Siegel found that elderly people who owned dogs were less likely to pay visits to their doctors after upsetting events than were those who didn’t 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 Group Dog Owners
12 8
10 5
6 9
9 4
15 6
12
14

a. Is there a significant difference in the number of doctor visits between dog owners and control subjects? Use a two-tailed test with á=.05.

You want to use a two-tailed t-test. You do not want to use matched pairs.

books set things up differently, but here is one way to do it..
group 1 = dog owners group 2 = no dog

Ho: mu1 - mu2 = 0
Ha: mu1 - mu2 does not = 0

If you are allowed to use a statistics calculator (like at TI-83 or 84) you can calculate very easily by entering each group into a different column and choosing 2 sample t-test. The be sure to choose the option not =. Compare the p-value given with the .05. If your value is less, reject the null hypothesis, if not fail to reject the null.

If you are not allowed to use a stat calculator, you will have to do this by using the formulas given in your text book.

Well, it looks like those dogs really are doing a good job of keeping the doctor away. Now let's see if the statistics agree with the canines.

To find out if there's a significant difference in the number of doctor visits between dog owners and control subjects, we can use a two-tailed test with alpha (á) set to 0.05.

First, let's organize the data a bit:

Control Group: 12, 10, 6, 9, 15, 12, 14
Dog Owners: 8, 5, 9, 4, 6

Hmm, it seems like we might have missed something in the Dog Owners group. There's a missing value there. But hey, dogs are known for their playful tendencies, so maybe one of them ran away with that data point!

Moving on, let's calculate the means for both groups:

Mean of Control Group: (12 + 10 + 6 + 9 + 15 + 12 + 14) / 7 = 11.57
Mean of Dog Owners: (8 + 5 + 9 + 4 + 6) / 5 = 6.4

Now let's find the standard deviations:

Standard Deviation of Control Group: 3.14
Standard Deviation of Dog Owners: 1.92

Since we have small sample sizes, we should use a t-test to compare the means. But before that, let's check the assumptions for this test. Are you ready? Here we go!

1. Independence: Well, we assume that each person's doctor visits are independent of others. Unless the doctor's office has turned into a party place, I don't see why this assumption wouldn't hold.

2. Normality: Here comes the tricky part. We need to check if the data distributions are normal. You could use some statistical tests or even plot the data to see if they look normal. However, since we only have a small sample size and the normality assumption is not that crucial for large sample sizes, let's just say it's not a big deal for now.

3. Homogeneity of Variance: We can check this by performing a test, such as Levene's Test. Or we can just cross our fingers and hope that the dog owners and control subjects have similar variances. Come on, let's cross those fingers!

Now, it's time for the t-test. We'll use the formula for an independent samples t-test:

t = (mean1 - mean2) / sqrt[(s1^2/n1) + (s2^2/n2)]

Plug in the numbers:

t = (11.57 - 6.4) / sqrt[(3.14^2/7) + (1.92^2/5)]

Calculating that, we get t ≈ 1.73.

Now, let's consult the magical table of critical values. With alpha (á) set to 0.05 for a two-tailed test, we need to find the critical values for t with degrees of freedom equal to 10 (7+5-2).

Checking the table, we see that for a two-tailed test at á = 0.05, the critical t-value is approximately 2.26.

Since 1.73 is less than 2.26, we fail to reject the null hypothesis. So it seems that there is no significant difference in the number of doctor visits between dog owners and control subjects, at least based on this small sample.

But hey, don't be disappointed. Remember, there are plenty of other benefits to owning dogs - like having a companion who can fetch your slippers or provide you with endless love and laughter. And who knows, they might even help you save on those doctor visits in the long run. Keep smiling!

To determine if there is a significant difference in the number of doctor visits between dog owners and control subjects, we can conduct a two-tailed t-test with an alpha level of 0.05.

Step 1: State the null hypothesis (H0) and alternative hypothesis (Ha):
Null hypothesis (H0): There is no significant difference in the number of doctor visits between dog owners and control subjects.
Alternative hypothesis (Ha): There is a significant difference in the number of doctor visits between dog owners and control subjects.

Step 2: Set the significance level (alpha):
Given alpha (á) = 0.05, which indicates a 5% chance of rejecting the null hypothesis when it is true.

Step 3: Find the degrees of freedom (df):
The degrees of freedom is calculated as (n1 - 1) + (n2 - 1), where n1 is the sample size of the control group and n2 is the sample size of the dog owners group. In this case, n1 = 5 and n2 = 7, so the degrees of freedom is 5 + 7 - 2 = 10.

Step 4: Calculate the test statistic:
For an independent two-sample t-test, the formula for the test statistic is:

t = (mean1 - mean2) / sqrt[(s1^2/n1) + (s2^2/n2)]

where mean1 and mean2 are the means of the control group and dog owners group respectively, s1 and s2 are the standard deviations of the control group and dog owners group respectively, and n1 and n2 are the sample sizes of the control group and dog owners group respectively.

Calculating the test statistic using the provided data:

Control Group:
Mean1 = (12 + 10 + 6 + 9 + 15) / 5 = 10.4
Standard Deviation1 = sqrt([(12-10.4)^2 + (10-10.4)^2 + (6-10.4)^2 + (9-10.4)^2 + (15-10.4)^2] / (5-1)) = 3.8

Dog Owner Group:
Mean2 = (8 + 5 + 9 + 4 + 6 + 12 + 14) / 7 = 8.7
Standard Deviation2 = sqrt([(8-8.7)^2 + (5-8.7)^2 + (9-8.7)^2 + (4-8.7)^2 + (6-8.7)^2 + (12-8.7)^2 + (14-8.7)^2] / (7-1)) = 3.9

Plugging the values into the test statistic formula:

t = (10.4 - 8.7) / sqrt[(3.8^2/5) + (3.9^2/7)] ≈ 0.511

Step 5: Calculate the p-value:
Using the calculated test statistic and the degrees of freedom, we can find the p-value associated with the t-value. From a t-table or statistical software, a two-tailed t-test with 10 degrees of freedom and alpha = 0.05 yields a p-value of approximately 0.623.

Step 6: Make a decision:
Compare the p-value (0.623) to the alpha level (0.05). Since the p-value (0.623) is greater than the alpha level (0.05), we fail to reject the null hypothesis.

Step 7: Interpret the result:
Based on the data provided, there is not enough evidence to conclude that there is a significant difference in the number of doctor visits between dog owners and control subjects at the 0.05 level of significance.

To determine if there is a significant difference in the number of doctor visits between dog owners and control subjects, we can conduct a hypothesis test.

First, we need to state our null and alternative hypotheses:

Null hypothesis (H0): There is no significant difference in the number of doctor visits between dog owners and control subjects.
Alternative hypothesis (Ha): There is a significant difference in the number of doctor visits between dog owners and control subjects.

Next, we can calculate the mean and standard deviation for the two groups:

Control Group:
Mean = (12 + 10 + 6 + 9 + 15 + 12 + 14) / 7 = 11.71
Standard Deviation = sqrt(((12-11.71)^2 + (10-11.71)^2 + (6-11.71)^2 + (9-11.71)^2 + (15-11.71)^2 + (12-11.71)^2 + (14-11.71)^2) / 6) = 2.702

Dog Owners:
Mean = (8 + 5 + 9 + 4 + 6) / 5 = 6.4
Standard Deviation = sqrt(((8-6.4)^2 + (5-6.4)^2 + (9-6.4)^2 + (4-6.4)^2 + (6-6.4)^2) / 4) = 1.624

Now, we can use the two-sample t-test to compare the means of the two groups. Since the sample sizes are small and we don't know the population standard deviations, we should use the t-distribution.

The t-test formula is:
t = (mean1 - mean2) / sqrt((sd1^2 / n1) + (sd2^2 / n2))

Plugging in the values:
t = (11.71 - 6.4) / sqrt((2.702^2 / 7) + (1.624^2 / 5)) = 1.87

Next, we need to determine the critical value or p-value to determine if the difference is significant.

Using a two-tailed test with a significance level of α = 0.05, and degrees of freedom equal to the sum of the sample sizes minus 2 (7 + 5 - 2 = 10), we can find the critical t-value.

By consulting a t-distribution table or using a statistical software, we find that the critical t-value for a two-tailed test with α = 0.05 and 10 degrees of freedom is approximately ±2.228.

Since 1.87 falls within the range of -2.228 to 2.228, it does not exceed the critical value. Therefore, we fail to reject the null hypothesis.

In conclusion, based on the given data and the two-tailed t-test with α = 0.05, there is not enough evidence to suggest a significant difference in the number of doctor visits between dog owners and control subjects.