Question

1. In the data sheet the results for two samples randomly taken from college and high school students SAT math scores. The standard deviations of both populations is unknown. Develop a 95% confidence interval for the difference between the two population means Is there conclusive evidence that one population has a larger mean? Explain.

Answer 1

Let $X_{1}$ be the random variable denoting the SAT scores of College student.

Let $X_{2}$ be the random variable denoting the SAT scores of High School student.

The 95% Confidence Interval is given by:

(T1 - C2 - s * tm1 +lz=2:0.025, T] - T2 + * tni+P2-2;0.025) 722 722


Now,

ni 16

$n_2=12$
$\overline{x_1}=\frac{\sum_{i=1}^{16}x_{1_i}}{16}=\frac{8400 }{16}=525$
$\overline{x_2}=\frac{\sum_{i=1}^{12}x_{2_i}}{12}=\frac{5844 }{12}=487$
$s_1=\sqrt{\frac{\sum_{i=1}^{16}(x_{1_i}-\overline{x_1})^2}{16-1}}=59.42053517$
$s_2=\sqrt{\frac{\sum_{i=1}^{12}(x_{2_i}-\overline{x_2})^2}{12-1}}=51.74763938$

Now,
$t_{12+16-2;0.025}=t_{26;0.025}=2.05553$

Putting in the given values, we get,
$\tiny (525 -487 -\sqrt{\frac{59.42053517 ^2}{16}+\frac{51.74763938 ^2}{12}}*2.05553,525 -487 +\sqrt{\frac{59.42053517 ^2}{16}+\frac{51.74763938 ^2}{12}}*2.05553)$

Hence, the 95% Confidence Interval is (-5.30,81.30) .

Now, since the Confidence Interval contains 0, there is no evidence that one population has a larger mean than the other.

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