To calculate the p-value for testing the difference in means between the two diets, we can use a two-sample t-test. First, we need to use the given data to compute the test statistic.
Given Data
-
Sample 1 (Diet 1):
- \( n_1 = 77 \)
- \( \bar{x}_1 = 4.7 \)
- \( s_1 = 7.16 \)
-
Sample 2 (Diet 2):
- \( n_2 = 79 \)
- \( \bar{x}_2 = 2.6 \)
- \( s_2 = 5.90 \)
Step 1: Calculate the pooled standard deviation (\(s_p\))
The pooled standard deviation is calculated using the formula:
\[
s_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}}
\]
Calculating each component:
- \(s_1^2 = 7.16^2 = 51.4256\)
- \(s_2^2 = 5.90^2 = 34.81\)
Now, substituting the values:
\[
s_p = \sqrt{\frac{(77 - 1)(51.4256) + (79 - 1)(34.81)}{(77 + 79 - 2)}}
\]
\[
= \sqrt{\frac{(76 \cdot 51.4256) + (78 \cdot 34.81)}{154}}
\]
Calculating the numerators:
\[
76 \cdot 51.4256 = 3908.7936
\]
\[
78 \cdot 34.81 = 2713.18
\]
Adding these:
\[
3908.7936 + 2713.18 = 6621.9736
\]
Now, the pooled standard deviation:
\[
s_p = \sqrt{\frac{6621.9736}{154}} \approx \sqrt{43.02068} \approx 6.63
\]
Step 2: Calculate the t-statistic
The t-statistic is calculated using the formula:
\[
t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}
\]
Substituting the values:
\[
t = \frac{4.7 - 2.6}{6.63 \sqrt{\frac{1}{77} + \frac{1}{79}}}
\]
\[
= \frac{2.1}{6.63 \sqrt{\frac{1.01299}{77}}}
\]
\[
= \frac{2.1}{6.63 \cdot 0.1131} \approx \frac{2.1}{0.74976} \approx 2.80
\]
Step 3: Degrees of Freedom
The degrees of freedom for the test is given by:
\[
df = n_1 + n_2 - 2 = 77 + 79 - 2 = 154
\]
Step 4: Find the p-value
Now we would look up the calculated \(t\) value in a t-distribution table or use a calculator/statistical software to find the p-value for \(|t| = 2.80\) with \(df = 154\).
Using software or an online calculator, you can find that:
\[
p \approx 0.0055 , (\text{two-tailed})
\]
Final Result
Thus, the p-value is approximately 0.0055.
