Parsimonious Peripherals has developed a new I / O device....

Parsimonious Peripherals has developed a new $I / O$ device. The company claims that the table contains the retrieval times of 30 randomly selected $I / O$ requests. The times are in milliseconds. Parsimonious claims the processing times are normally distributed with a mean of 10 milliseconds and a standard deviation of 5 milliseconds. Test this claim using the chi-square test as follows. Normalize the data by the formula $z_i=\left(x_i-10\right) / 5$. Partition the real line into 5 intervals such that if $H_0$ is true, then one-fifth of the $z_i^{\prime} s$ would be expected in each interval. Apply the chi-square test with these intervals as cells. Use the $5 \%$ level of significance. $$ \begin{array}{rrrrr} {\text { Parsimonious Sample }} \\ \hline & & & & \\ 14.560 & 4.102 & 7.514 & 8.873 & 10.373 \\ 7.582 & 10.651 & 5.424 & 6.470 & 7.589 \\ 3.022 & 8.666 & 1.966 & 17.381 & 15.564 \\ 17.715 & 13.655 & 14.860 & 12.037 & 10.475 \\ 5.298 & 9.750 & -1.371 & 7.274 & 14.901 \\ 18.845 & 16.881 & 7.229 & 16.833 & 9.440 \\ \hline \end{array} $$

Parsimonious Peripherals has developed a new $I / O$ device. The company claims that the table contains the retrieval times of 30 randomly selected $I / O$ requests. The times are in milliseconds. Parsimonious claims the processing times are normally distributed with a mean of 10 milliseconds and a standard deviation of 5 milliseconds. Test this claim using the chi-square test as follows. Normalize the data by the formula $z_i=\left(x_i-10\right) / 5$. Partition the real line into 5 intervals such that if $H_0$ is true, then one-fifth of the $z_i^{\prime} s$ would be expected in each interval. Apply the chi-square test with these intervals as cells. Use the $5 \%$ level of significance.
$$
\begin{array}{rrrrr}
{\text { Parsimonious Sample }} \\
\hline & & & & \\
14.560 & 4.102 & 7.514 & 8.873 & 10.373 \\
7.582 & 10.651 & 5.424 & 6.470 & 7.589 \\
3.022 & 8.666 & 1.966 & 17.381 & 15.564 \\
17.715 & 13.655 & 14.860 & 12.037 & 10.475 \\
5.298 & 9.750 & -1.371 & 7.274 & 14.901 \\
18.845 & 16.881 & 7.229 & 16.833 & 9.440 \\
\hline
\end{array}
$$

Key Concepts

Significance Level

The significance level, often denoted by alpha, represents the threshold at which the null hypothesis is rejected. It is the probability of committing a Type I error—rejecting the null hypothesis when it is actually true. In hypothesis testing, comparing the p-value or the test statistic to the critical value determined by the significance level (such as 5%) provides a rigorous criterion for deciding whether the observed data is consistent with the stated hypothesis.

Equal Probability Intervals

Partitioning the range of a distribution into intervals such that each interval has the same probability of occurrence under the null hypothesis is a common step in the chi-square goodness-of-fit test. By dividing the standardized data into cells or bins with equal theoretical probabilities, one can accurately calculate the expected frequency for each cell. This method ensures that any deviation between the observed and expected counts is due to the sample data rather than uneven interval widths.

Chi-Square Goodness-of-Fit Test

This is a statistical method used to compare an observed frequency distribution with an expected frequency distribution under a specific hypothesis. In the context of testing whether a sample comes from a given theoretical distribution, the test calculates a statistic by summing the squared differences between observed and expected counts in each category, normalized by the expected counts. The resulting statistic is then compared with a critical value from the chi-square distribution to decide whether to reject the null hypothesis that the observed data fits the expected distribution.

Normal Distribution

A normal distribution is a continuous probability distribution characterized by a symmetrical bell-shaped curve, defined by its mean and standard deviation. Many natural phenomena and measurement errors tend to follow this distribution due to the central limit theorem. When testing hypotheses about data, assuming normality allows researchers to make probabilistic statements and apply various statistical tests that depend on the distribution’s properties.

Data Normalization

Data normalization, in this context, involves transforming raw observations into standardized values by subtracting the mean and dividing by the standard deviation. This procedure converts the data into z-scores, which then follow a standard normal distribution (with mean 0 and standard deviation 1) under the null hypothesis. Normalization simplifies calculations and comparisons by removing the effect of the original scale of measurement.

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