Reference no: EM131733614
Question: 1. A research was interested to see if there is a difference between male and female on reading squinting. Here are the outputs.
Gender: Male , Female
N: 48 (Male) 161 (Female)
Mean: .42 (Male) .57 (Female)
Std.: .577 (Male) .650 (Female)
Std Error Mean: .083 (Male) .051 (Female)
Independent Samples Test
Levene's test for equality for variance
Equality of variance assumed:
F: .968
Sig: .326
t-test for equality of means
Equality of variance assumed:
t: -1.425
Sig. (2 tailed): .156
95% Confidence interval of difference : Lower ( -.354 ) Upper ( .057)
Equality of variance not assumed
t:-1.519
df: 85.620
Sig (2 tailed): .133
95% Confidence interval of difference : Lower ( -.343 ) Upper ( .046)
a. State the null and alternative hypothesis.
b. Calculate the degrees of freedom
c. What is the p value of the test?
d. What is the test statistic value?
e. Are you going to accept or reject the null hypothesis?
f. What is your conclusion? If you have rejected the null, what is the effect size?
2. Dr. Smith investigated to see if there is a difference in headaches between the individuals who wear corrective lenses and those who don't. Below are the outputs:
Group Statistics
Headaches
No N: 94 Mean: .60 Std. : .555 Std Error Mean: .057
Yes N: 115 Mean: .57 Std. : .515 Std Error Mean: .048
Independent Sample Test
Levene's test for equality for variance
Equality of variance assumed:
F: .784
Sig: .377
t-test for equality of means
Equality of variance assumed:
t: .412
Sig. (2 tailed): .681
95% Confidence interval of difference : Lower ( -.116 ) Upper ( .177)
Equality of variance not assumed
t:-.409
Sig (2 tailed): .683
95% Confidence interval of difference : Lower ( -.117 ) Upper ( .178)
a. State the null and alternative hypothesis.
b. Calculate the degrees of freedom
c. What is the p value of the test?
d. What is the test statistic value?
e. Are you going to accept or reject the null hypothesis?
f. What is your conclusion? If you have rejected the null, what is the effect size?
3. A researcher in exercise physiology wanted to know whether body composition differs among prepubescent males and females. To test the null hypothesis, she measured skinfolds on 5 males and 6 females and obtained the following results:
mean for males: 25, SD= 2.35
Mean for females: 21, SD= 3.64
Pooled SD= 3.84
Alpha level: .05
a. What is the obtained value?
b. Calculate the degrees of freedom?
c. What is the result?