Dear beloved readers, welcome to our website! We hope your visit here brings you valuable insights and meaningful inspiration. Thank you for taking the time to stop by and explore the content we've prepared for you.
------------------------------------------------ A researcher was interested in whether the cranial capacity of one species of primates was larger than the cranial capacity of another species. She collected 8 random skulls from species 1 and 9 random skulls from species 2. These were the cranial capacities of the skulls in mm$^3$:

Species 1:
\[ c(92.0, 157.2, 104.0, 99.8, 102.8, 176.2, 64.0, 113.4) \]

Species 2:
\[ c(40.9, 127.3, 101.8, 147.2, 50.5, 75.7, 71.8, 121.5, 86.8) \]

- What are the degrees of freedom?
- What is the observed t-value?
- What is the lower bound of the 95% Confidence Interval for the difference in means?
- What is the p-value?
- Does this test suggest that the population mean of cranial capacities in species 1 is larger than species 2?

d) The p-value for a one-tailed t-test with 15 degrees of freedom and a t-value of 1.89 can be found using a t-distribution table or a statistical software.

e) The p-value of 0.04 suggests that there is a 4% chance of obtaining a difference in means as extreme or more extreme than the observed difference, assuming that the null hypothesis is true.

The degrees of freedom are the number of independent observations in a statistical analysis that can vary freely. In this scenario, the degrees of freedom for the t-test can be calculated using the following formula:

df = n1 + n2 - 2

where n1 is the sample size of species 1 (8) and n2 is the sample size of species 2 (9).

df = 8 + 9 - 2 = 15

The observed t-value is a measure of how different the sample means are from each other in standard error units. It can be calculated using the following formula:

t = (x1 - x2) / SE

where x1 is the sample mean of species 1 (121.1), x2 is the sample mean of species 2 (89.1), and SE is the standard error of the difference in means. The formula for the standard error is:

SE = √[(s1² / n1) + (s2² / n2)]

where s1 is the sample standard deviation of species 1 (37.9), s2 is the sample standard deviation of species 2 (30.7), n1 is the sample size of species 1 (8), and n2 is the sample size of species 2 (9).

SE = √[(37.9² / 8) + (30.7² / 9)] = 16.9

Substituting these values into the formula for t:

t = (121.1 - 89.1) / 16.9 = 1.89

The 95% confidence interval is a range of values within which we can be 95% confident that the true population mean lies. The lower bound of the 95% confidence interval can be calculated using the following formula:

Lower bound = (x1 - x2) - t(alpha/2) * SE

where x1 is the sample mean of species 1 (121.1), x2 is the sample mean of species 2 (89.1), t(alpha/2) is the t-value for the given alpha level (0.025 for a two-tailed test at 95% confidence), and SE is the standard error of the difference in means calculated above.

Lower bound = (121.1 - 89.1) - 2.131 * 16.9 = 12.9

The t-value for the null hypothesis can be calculated using the same formula as before, but with the difference in means set to zero:

t_null = (x1 - x2 - 0) / SE = (121.1 - 89.1) / 16.9 = 1.89

The p-value is the probability of getting a t-value as extreme or more extreme than 1.89 under the null hypothesis. From a t-distribution table, we can see that the p-value is approximately 0.04 for a one-tailed test.

Since the p-value is less than the conventional significance level of 0.05, we reject the null hypothesis and conclude that there is evidence to suggest that the population mean cranial capacity of species 1 is larger than that of species 2.

To know more about p-value here

https://brainly.com/question/14790912

#SPJ4