Question

Two researchers conducted a study in which two groups of students were asked to answer 42 trivia questions from a board game. The students in group 1 were asked to spend 5 minutes thinking about what it would mean to be a professor, while the students in group 2 were asked to think about soccer hooligans. These pretest thoughts are a form of priming. The 200 students in group 1 had a mean score of 21.4 with a standard deviation of 3.6, while the 200 students in group 2 had a mean score of 19.2 with a standard deviation of 4.2. Complete parts (a) and (b) below. (a) Determine the 95% confidence interval for the difference in scores, , -H2. Interpret the interval. The lower bound is The upper bound is (Round to three decimal places as needed.) Interpret the interval. Choose the correct answer below. O A. There is a 95% probability that the difference of the means is in the interval. O B. The researchers are 95% confident that the difference of the means is in the interval. O C. The researchers are 95% confident that the difference between randomly selected individuals will be in the interval. OD. There is a 95% probability that the difference between randomly selected individuals will be in the interval. (b) What does this say about priming? O A. Since the 95% confidence interval contains zero, the results suggest that priming does not have an effect on scores. O B. Since the 95% confidence interval does not contain zero, the results suggest that priming does not have an effect on scores. O C. Since the 95% confidence interval contains zero, the results suggest that priming does have an effect on scores. OD. Since the 95% confidence interval does not contain zero, the results suggest that priming does have an effect on scores

Answer 1

GIVEN:

Sample size of group 1

n1) = 200

Sample mean score of group 1

(1)= 21.4

Sample standard deviation of group 1

813.6

Sample size of group 2

(п2) 200 T

Sample mean score of group 2

2) = 19.2

Sample standard deviation of group 2 $(s_{2})=4.2$

(a) 95% CONFIDENCE INTERVAL FOR DIFFERENCE IN MEAN SCORES:

HYPOTHESIS:

The hypothesis is given by,

$H_{0}:\mu _{1}-\mu _{2}=0$

HI 2 0

FORMULA USED:

The 95% confidence interval for difference in mean scores $(\mu _{1}-\mu _{2})$ is,

$(\bar{x_{1}}-\bar{x_{2}})\pm t_{c}*\sqrt{[(s_{1}^2/n_{1})+(s_{2}^2/n_{2})]}$

where $t_{c}$ is the t critical value at 95% confidence level with degrees of freedom given by,

$df=[(s_{1}^2/n_{1})+(s_{2}^2/n_{2})]^2/[[(s_{1}^2/n_{1})^2/(n_{1}-1)]+[(s_{2}^2/n_{2})^2/(n_{2}-1)]]$

DEGREES OF FREEDOM:

The degrees of freedom given by,

$df=[(s_{1}^2/n_{1})+(s_{2}^2/n_{2})]^2/[[(s_{1}^2/n_{1})^2/(n_{1}-1)]+[(s_{2}^2/n_{2})^2/(n_{2}-1)]]$

$=[(3.6^2/200)+(4.2^2/200)]^2/[[(3.6^2/200)^2/(200-1)]+[(4.2^2/200)^2/(200-1)]]$

$=0.0234/[0.000021+0.000039]$

$df=390$

CRITICAL VALUE:

The two tailed t critical value with 390 degrees of freedom at 95% confidence level is .

CALCULATION:

The 95% confidence interval for difference in mean scores $(\mu _{1}-\mu _{2})$ is,

$(\bar{x_{1}}-\bar{x_{2}})\pm t_{c}*\sqrt{[(s_{1}^2/n_{1})+(s_{2}^2/n_{2})]}$

  $=(21.4-19.2)\pm1.9661*\sqrt{[(3.6^2/200)+(4.2^2/200)]}$

  

  $=[1.431,2.969]$

The 95% confidence interval for difference in mean scores $(\mu _{1}-\mu _{2})$ is .

INTERPRETATION:

The correct answer is Option (B): The researchers are 95% confident that the difference of the means is in the interval.

(b) CONCLUSION:

The correct answer is Option (D): Since the 95% confidence interval does not contain zero, the result suggest that priming does have an effect on scores.

8) GIVEN:

Sample size of adults with no children $(n_{1})=40$

Sample mean daily leisure time of adults with no children $(\bar{x_{1}})=5.27$ hours

Sample standard deviation of of adults with no children $(s_{1})=2.47$ hours

Sample size of adults with children $(n_{2})=40$

Sample mean daily leisure time of adults with children $(\bar{x_{2}})=4.36$ hours

Sample standard deviation of of adults with children $(s_{2})=1.75$ hours

(a) 95% CONFIDENCE INTERVAL FOR MEAN DIFFERENCE IN LEISURE TIME BETWEEN ADULTS WITH NO CHILDREN AND ADULTS WITH CHILDREN:

HYPOTHESIS:

The hypothesis is given by,

$H_{0}:\mu _{1}-\mu _{2}=0$

HI 2 0

FORMULA USED:

The 95% confidence interval for mean difference in leisure time between adults with no children and adults with children $(\mu _{1}-\mu _{2})$ is,

$(\bar{x_{1}}-\bar{x_{2}})\pm t_{c}*\sqrt{[(s_{1}^2/n_{1})+(s_{2}^2/n_{2})]}$

where $t_{c}$ is the t critical value at 95% confidence level with degrees of freedom given by,

$df=[(s_{1}^2/n_{1})+(s_{2}^2/n_{2})]^2/[[(s_{1}^2/n_{1})^2/(n_{1}-1)]+[(s_{2}^2/n_{2})^2/(n_{2}-1)]]$

DEGREES OF FREEDOM:

​​​​​​​The degrees of freedom given by,

$df=[(s_{1}^2/n_{1})+(s_{2}^2/n_{2})]^2/[[(s_{1}^2/n_{1})^2/(n_{1}-1)]+[(s_{2}^2/n_{2})^2/(n_{2}-1)]]$

  $=[(2.47^2/40)+(1.75^2/40)]^2/[[(2.47^2/40)^2/(40-1)]+[(1.75^2/40)^2/(40-1)]]$

  $=0.0525/[0.000596+0.00015]$

$df=70.4\approx 70$

CRITICAL VALUE:

The two tailed t critical value with 70 degrees of freedom at 95% confidence level is .

CALCULATION:

The 95% confidence interval for mean difference in leisure time between adults with no children and adults with children $(\mu _{1}-\mu _{2})$ is,

$(\bar{x_{1}}-\bar{x_{2}})\pm t_{c}*\sqrt{[(s_{1}^2/n_{1})+(s_{2}^2/n_{2})]}$

  $=(5.27-4.36)\pm1.9944*\sqrt{[(2.47^2/40)+(1.75^2/40)]}$

  

  $=[-0.04,1.86]$

The 95% confidence interval for $(\mu _{1}-\mu _{2})$ is .

INTERPRETATION:

Since the value of the parameter specified by the null hypothesis (0) is contained in the 95% interval , the null hypothesis cannot be rejected at the 0.05 level. Thus the correct answer is Option (D): There is 95% confidence that the difference of the means is in the interval. Conclude that there is insufficient evidence of a significant difference in the number of leisure hours.