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Understanding expected frequencies is crucial when conducting a chi-square test. The concept of expected frequencies relates to what we would anticipate the results to look like if theoretical assumptions are true. In the context of our nut type example, expected frequencies are predictions based on the percentages provided.

For a sample size of 200 nuts, with types supposed to be distributed as 40%, 30%, 20%, and 10%, the expected frequencies are calculated by multiplying each percentage by the sample size. This straightforward multiplication gives us a benchmark to compare the actual observed results against. If the observed frequencies divert significantly from these expected counts, it suggests that the actual distribution might not align with our expectations.

Mathematically, the calculation formula would be given by: \[Expected \ Frequency_{Type_i} = Sample \ Size \times Percentage_{Type_i}/100\] for each type of nut, where i ranges from 1 to 4. These figures are the cornerstone for determining whether the observed frequencies result from random chance or indicate a more systemic discrepancy.

Statistical significance acts as a flag to signal whether the findings from our statistical test are likely to be due to a genuine effect, rather than just random chance. Essentially, it helps us make decisions based on the data.

In the chi-square test, we compare the calculated test statistic to a critical value from the chi-square distribution table. This critical value depends on the chosen level of significance (commonly \( 0.05 \), \( 0.01 \), or \( 0.001 \)) and the degrees of freedom (which depends on the number of categories minus one in our case).

If our calculated test statistic exceeds the critical value, we consider the results to be statistically significant and thus, we have enough evidence to reject the null hypothesis. In our example, with a test statistic of \(X^2 = 19.0\) and a significance level of \(0.001\), the conclusion was to reject the null hypothesis as the value exceeded the critical threshold.

The null hypothesis in a chi-square test typically states that there is no significant difference between the expected and observed frequencies. It serves as a default position that suggests no association or effect.

For our nut types example, the null hypothesis would state that the sample distribution of various nut types perfectly reflects the claimed percentages of 40%, 30%, 20%, and 10%. When we perform the chi-square test, we're essentially putting this hypothesis to test.

If our calculated chi-square statistic is less than the critical value, we do not have enough evidence to dismiss this default assumption. However, in our exercise, the test statistic was 19.0, which is above the critical value for a significance level of \(0.001\), leading to a rejection of the null hypothesis - suggesting that the actual distribution of nuts may differ from what's claimed.

The sampling distribution is a probability distribution of a statistic that arises from a large number of samples of the same size from a certain population. In simpler terms, it tells us how the results of an experiment would behave if we were to repeat it many times.

This distribution is key in estimating probabilities and making inferences about the population from which the samples are drawn. For the chi-square test, we assume the sample size should be large enough so the resulting sampling distribution of the test statistic can be approximated by a chi-square distribution.

In cases where the sample size is small - for instance, a sample size of 40 as proposed in part b of our exercise - some expected frequencies become less than 5, which invalidates the normal approximation. Therefore, it would not be advisable to use the chi-square test for such small samples, as it could lead to inaccurate conclusions.