Answer:
From the given data, there is not enough evidence to prove that there is a statistically significant difference between the two population means
Step-by-step explanation:
The number of younger moms, in the study = 837
The average weight gain of younger moms,
₁ = 30.67 pounds
The standard deviation of the weight gain of younger moms, s₁ = 14.69 pounds
(30.67 - 28.52)/√((14.69^2)/837 + 13²/143))
The number of younger moms, in the study = 143
The average weight gain of mature moms,
₂ = 28.52 pounds
The standard deviation of the weight gain of mature moms, s₂ = 13 pounds
The test statistic for the difference in two populations is given as follows;
![t=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{\dfrac{s_1^{2} }{n_{1}}-\dfrac{s _{2}^{2}}{n_{2}}}}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7B%28%5Cbar%7Bx%7D_%7B1%7D-%5Cbar%7Bx%7D_%7B2%7D%29%7D%7B%5Csqrt%7B%5Cdfrac%7Bs_1%5E%7B2%7D%20%7D%7Bn_%7B1%7D%7D-%5Cdfrac%7Bs%20_%7B2%7D%5E%7B2%7D%7D%7Bn_%7B2%7D%7D%7D%7D)
Therefore, we get;
![t=\dfrac{(30.67-28.52)}{\sqrt{\dfrac{14.69^{2} }{837}-\dfrac{13^{2}}{143}}} \approx 1.7919](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7B%2830.67-28.52%29%7D%7B%5Csqrt%7B%5Cdfrac%7B14.69%5E%7B2%7D%20%7D%7B837%7D-%5Cdfrac%7B13%5E%7B2%7D%7D%7B143%7D%7D%7D%20%5Capprox%201.7919)
The test statistic ≈ (1.7919)
Using a graphing calculator, we get;
The critical-t = ±1.971379, p = 0.07459697
Therefore, given that the test statistic, (1.7919), < critical-t (0.07459697), we fail to reject the null hypothesis, therefor, the given data does not provide convincing evidence that there is a significant difference between the two population means