Test Statistic Formula:
?
=
?
ˉ
1
−
?
ˉ
2
?
1
2
?
1
+
?
2
2
?
2
t=
n
1

s
1
2

+
n
2

s
2
2

x
ˉ
1

−
x
ˉ
2

Substitute:

?
=
23.6
−
20.1
4.8
2
50
+
5.0
2
55
t=
50
4

Question

Test Statistic Formula:
?
=
?
ˉ
1
−
?
ˉ
2
?
1
2
?
1
+
?
2
2
?
2
t=
n
1
	​

s
1
2
	​

​

+
n
2
	​

s
2
2
	​

​

x
ˉ
1
	​

−
x
ˉ
2
	​

​

Substitute:

?
=
23.6
−
20.1
4.8
2
50
+
5.0
2
55
t=
50
4.8
2
	​

+
55
5.0
2
	​

​

23.6−20.1
	​

=
3.5
23.04
50
+
25
55
=
50
23.04
	​

+
55
25
	​

​

3.5
	​

=
3.5
0.4608
+
0.4545
=
3.5
0.9153
=
0.4608+0.4545
	​

3.5
	​

=
0.9153
	​

3.5
	​

?
=
3.5
0.9567
≈
3.66
t=
0.9567
3.5
	​

≈3.66
Step 3 — Degrees of Freedom (Welch–Satterthwaite)
?
?
=
(
?
1
2
?
1
+
?
2
2
?
2
)
2
(
?
1
2
?
1
)
2
?
1
−
1
+
(
?
2
2
?
2
)
2
?
2
−
1
df=
n
1
	​

−1
(
n
1
	​

s
1
2
	​

​

)
2
	​

+
n
2
	​

−1
(
n
2
	​

s
2
2
	​

​

)
2
	​

(
n
1
	​

s
1
2
	​

​

+
n
2
	​

s
2
2
	​

​

)
2
	​

?
?
=
(
0.4608
+
0.4545
)
2
(
0.4608
)
2
49
+
(
0.4545
)
2
54
df=
49
(0.4608)
2
	​

+
54
(0.4545)
2
	​

(0.4608+0.4545)
2
	​

?
?
=
(
0.9153
)
2
0.2123
49
+
0.2066
54
≈
0.8378
0.00816
≈
102.68
df=
49
0.2123
	​

+
54
0.2066
	​

(0.9153)
2
	​

≈
0.00816
0.8378
	​

≈102.68
?
?
≈
103
df≈103
Step 4 — P-value

Now,

?
=
3.66
,
?
?
=
103
t=3.66,df=103

For a one-tailed test:

?
=
?
(
?
103
>
3.66
)
p=P(T
103
	​

>3.66)

Using Excel:

=
1
−
?
.
?
?
?
?
(
3.66
,
103
,
?
?
?
?
)
=1−T.DIST(

VIPSolutions · Accepted Answer

# Hypothesis Test Calculation

## Test Statistic Formula

$$
t = \frac{\bar{x1 - \bar{x}_2}{\sqrt{\frac{s_12}{n_1} + \frac{s_^2}{_2}}$$

---

## Step 1 Substituting

$$
 = \frac{23.6 20.1}{\sqrtfrac{4.8^250} + \frac{5.2}{55}}}
$$

Calculate numerator:

$$
23.6 - 20.1 = 3.5
$$

Calculate denominator:

$$
\sqrt{\frac{23.04}{50} + \frac{25}{55}} = \sqrt{.4608 + .4545} = \sqrt{.9153} \approx .9567
$$

Compute $ t $:

$$
t = \frac{3.5}{.9567} \approx 3.66
$$

---

## Step 2 — Degrees of Freedom (Welch–Satterthwaite)

$$
df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}}
$$

Calculate numerator:

$$
(.4608 + .4545)^2 = (.9153)^2 \approx .8378
$$

Calculate denominator:

$$
\frac{(.4608)^2}{49} + \frac{(.4545)^2}{54} = \frac{.2123}{49} + \frac{.2066}{54} \approx .00434 + .00382 = .00816
$$

Compute degrees of freedom:

$$
df = \frac{.8378}{.00816} \approx 102.68 \approx 103
$$

---

## Step 3 — P-value Calculation

Given:

$$
t = 3.66, \quad df = 103
$$

For a one-tailed test:

$$
p = P(T_{103} > 3.66)
$$

Using Excel:

```excel
= 1 - T.DIST(3.66, 103, TRUE)
```

Result:

$$
p \approx .0002
$$

---

## Step 4 — Decision and Conclusion

Since:

$$
p = .0002 < .05
$$

**Reject $H_$.**

The evidence suggests a statistically significant difference between the groups.

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