Question

a researcher wants to investigate whether there is a difference in the aerobic capacity between two different exercise programs. group 1 participates in a running program, and group 2 participates in a cycling program. after 8 weeks, the aerobic capacity is measured for both groups. below are the sample data:
group 1 (running program): sample size ($n_1$) = 3, sample mean ($overline{x}_1$) = 40, sample variance ($s_1^2$) = 4
group 2 (cycling program): sample size ($n_2$) = 4, sample mean ($overline{x}_2$) = 45, sample variance ($s_2^2$) = 9.
question 2:
what is the p - value?
.056
.048
.028
.007

Explanation:

Step1: Calculate the t - statistic

First, we assume equal variances. The pooled variance \(s_p^2=\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}\)
\(n_1 = 3\), \(s_1^2=4\), \(n_2 = 4\), \(s_2^2 = 9\)
\(s_p^2=\frac{(3 - 1)\times4+(4 - 1)\times9}{3 + 4-2}=\frac{8 + 27}{5}=\frac{35}{5}=7\)

The standard error \(SE=\sqrt{s_p^2(\frac{1}{n_1}+\frac{1}{n_2})}=\sqrt{7(\frac{1}{3}+\frac{1}{4})}=\sqrt{7\times\frac{4 + 3}{12}}=\sqrt{\frac{49}{12}}\approx2.02\)

The t - statistic \(t=\frac{\bar{X}_1-\bar{X}_2}{SE}\), \(\bar{X}_1 = 40\), \(\bar{X}_2=45\)
\(t=\frac{40 - 45}{2.02}\approx - 2.48\)

Step2: Determine the degrees of freedom

The degrees of freedom \(df=n_1 + n_2-2=3 + 4-2 = 5\)

Step3: Find the p - value

Using a t - distribution table or a statistical software for a two - tailed test with \(t=-2.48\) and \(df = 5\)
\(P(|t|>2.48)\)
Looking up in the t - table: \(P(t>2.48)\) for \(df = 5\) is between \(0.025\) and \(0.05\)
Using a calculator or software (e.g., in R: \(2*(1 - pt(2.48,5))\approx0.056\))

Answer:

0.056