Step 1

Independent Two Samples t-test:

Suppose there are two populations independent of one another. A researcher is interested to know whether the populations are equal on an average, or in other words, whether the means of the two populations are equal or not, based on sample observations.

Suppose \(n_{x}\) random observations,

\(X_{1}, X_{2},...X_{n}\) are taken from the first population and

\(n_{y}\) random observations,

\(Y_{1}, Y_{2},...Y_{n}\) are taken from the second population.

Let \(\mu_{x}, \mu_{y}\) be the true population means of the two populations, respectively. Further, assume that the population standard deviations of the two populations \(\sigma_{x}, \sigma_{y}\) are unknown and must be estimated using the corresponding sample standard deviations, \(s_{x}, s_{y}.\)

In such a situation, the comparison between the two population means is done using the Independent Two Samples t-test, provided the following assumptions are satisfied:

1) The observations should be from a continuous data.

2) The populations should be independent of one another.

3) The samples must be obtained randomly, so that the observations are independent of each other for each population.

4) The data must follow the normal distribution, or the sample sizes must be large enough to ensure an approximate normal distribution.

Ideally, an independent t-test would require two sets of independent individuals, one set of individuals are 8-year-old children and the other set of individuals are 10-year-old children. Here, the vocabulary skills of both the teams are two independent samples. To test the difference between two independent samples, independent t-test would be appropriate.

Thus, independent sample t-test is recommended.

Therefore, Independent t-test would be used to determine if there is a significant difference in the vocabulary skills between 8-year-old and 10-year-old.

Step 2

Null hypothesis \(H_{0}:\) There is no significant difference in the mean vocabulary skills between the 8-year-old and 10-year-old.

Alternative hypothesis \(H_{0}:\)

There is a significant difference in the mean vocabulary skills between the 8-year-old and 10-year-old.

The sample statistic is given below:

\(\overline{x}_{1}=\) Mean vocabulary skill rating of 8-year-old

\(\overline{x}_{2}=\) Mean vocabulary skill rating of 10-year-old

sample statistic \(= (\overline{x}_{1} - \overline{x}_{2})\)

The test statistic is given below:

Test statistic \(=\frac{(\overline{x}_{1}-\overline{x}_{2})-(\mu_{1}-\mu_{2})}{\sqrt{S_{p}^{2}(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}\)

\(S_{p}^{2} =\) Pooled variance asuming equal population variances

The standard error is given below:

\(S_{e} =\sqrt{S_{p}^{2}(\frac{1}{n_{1}}+\frac{1}{n_{2}})}\)

Level of significance:

Type-I error is rejecting the null hypothesis \(H_{0}\),

when the null hypothesis \(H_{0}\) is true.

The probability of type-I error is level of significance. Therefore, the level of significance is the probability of making a mistake.