Question

The intelligence quotients (IQs) of 16 students from one area of a city showed a mean of 107 and a standard deviation of 10, while IQ of 14 students from another area of the city showed a mean of 112 and a standard deviation of 8. Is there a significant difference between IQs of the two groups at significance level of (a) 0.01 and (b) 0.05?

Answer 1

Answer on Question #78963 – Math – Statistics and Probability

Question

The intelligence quotients (IQs) of 16 students from one area of a city showed a mean of 107 and a standard deviation of 10, while IQ of 14 students from another area of the city showed a mean of 112 and a standard deviation of 8. Is there a significant difference between IQs of the two groups at significance level of (a) 0.01 and (b) 0.05?

Solution

Performing independent samples t-test, not assuming equal variances.

Assumptions: both populations must be normal.

The null hypothesis: the mean IQs are equal.

The alternative hypothesis: the mean IQs are different.

Degrees of freedom: $df = \min(N1, N2) - 1 = 13$

The standard error:

$$SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{10^2}{16} + \frac{8^2}{14}} = 3.2896$$

The test statistics:

$$t = \frac{\left(\bar{x}_1 - \bar{x}_2\right) - 0}{SE} = \frac{107 - 112}{3.2896} = -1.52$$

The two-tailed cumulative probability value associated with the given $t$-statistic can be determined from the Student's $t$-distribution table or calculated using the technology (function T.DIST.2T() of MS Excel).

For $df = 13$ and $t = -1.52$, $p = 0.152$

Since the $p$-value is greater than both $\alpha$ values, fail to reject the null hypothesis at both significance levels.

The samples do not provide sufficient evidence to conclude the difference between the mean IQs.

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