Software Lab 4.2 Solutions

Lesson 4.2: Testing Goodness of Fit in One-way Tables

The expected frequencies are each 100, so they are all at least five.
$\chi^2 = \dfrac{(96-100)^2}{100} + \dfrac{(94-100)^2}{100} + \dfrac{(90-100)^2}{100} +$ $\dfrac{(89-100)^2}{100} + \dfrac{(114-100)^2}{100} + \dfrac{(117-100)^2}{100} = 7.58$ .
1 - pchisq(7.58, df=5) ≈ 0.181. Since p-value $= 0.1810 > \alpha = 0.05$ , there is insufficient evidence to reject the null hypothesis. The data support the claim that the die is fair.
The test statistic and p-value match:

Figure 1: Chi-square goodness -of-fit test
The expected frequencies are 742, 338, 154, 70, 32, and 14.
The expected frequency for 7+ days is 12.
The expected frequencies are all at least five.
$\chi^2 = \dfrac{(717-742)^2}{742} + \dfrac{(369-338)^2}{338} + \dfrac{(155-154)^2}{154} +$ $\dfrac{(69-70)^2}{70} + \dfrac{(28-32)^2}{32} + \dfrac{(14-14)^2}{14} + \dfrac{(10-12)^2}{12} = 4.54$ .
1 - pchisq(4.54, df=6) ≈ 0.604. Since p-value $= 0.604 > \alpha = 0.05$ , there is insufficient evidence to reject the null hypothesis. The data support the claim that trading days are independent.
The test statistic and p-value match:

Figure 2: Chi-square goodness-of-fit test