It is to be noted that the determination of whether or not there is a significant difference between the two groups will be done using a t test.
A t-test is a statistical test that juxtaposes two samples' means. It is used in hypothesis testing, using a null hypothesis that the variance in group means is zero and an alternative hypothesis that the difference is not zero.
The conditions to use the t test are:
The degrees of freedom (k) to be utilized for this text will be derived using the conservative method given below:
df = [(s₁²/n₁) + (s₂²/n²)/[((s₁²/n₁)²/((n₁-1)) + (s₂²/n₂)²/((n₂-1))]
= [(3.0952/15) + (6.4095/15)]² / [((3.0952/15)²/14) + ((6.4095/15)²/14)]
= 24.965
Hence,
df ≈ 24 (if approximated to the floor)
Recall the the standard deviation of the population are unequal and unknown. This thus requires that we utilize the two-sample unpooled t-test.
Here, H₀ is given as;
[tex]t = \frac{\bar{x_{1} -\bar{x_{2}}}}{\sqrt{\frac{s_{1}^{2} }{n_{1} } + \frac{S_{2}^{2} }{n_{2}} } } \sim t_{df}[/tex]
t = [(3.33333 - 4.13333)]/√[(3.0952/15) + (6.4095/15)]
= - 0.8/√0.6337
t = - 1.005
The 95% confidence interval is computed using the following formula:
[tex](\bar x_{2} - \bar x_{1}) \pm t_\alpha_/_2_,_df \left({\sqrt{\frac{S_{1}^{2} }{n_{1} } + \frac{S_{2}^{2} }{n_{2}} } } \right)[/tex]
= - 0.8 ± t₀.₀₂₅,₂₄ (√0.6337)
= - 0.8 ± 2.064 (√0.6337)
= -2.4429, 0.8429
To derive the 90% interval, we state:
[tex](\bar x_{2} - \bar x_{1}) \pm t_\alpha_/_2_,_df \left({\sqrt{\frac{S_{1}^{2} }{n_{1} } + \frac{S_{2}^{2} }{n_{2}} } } \right)[/tex]
= - 0.8 ± t₀.₀₅₀,₂₄ (√0.6337)
= - 0.8 ± 0.685 (√0.6337)
= -2.162, 0.562
From the intervals computed, we must fail to reject H₀
H₀ : μ₁ = μ₂
It is clear from the above intervals computed from that the differences between the mean of both groups is significant. This is because, zero is included on the two intervals.
Learn more about t-test at;
https://brainly.com/question/6589776
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