Question

1. Consider again the situation described in Exercise 11 of Sec. 9.6. Test the null hypothesis that the variance of the fusion time for subjects who saw a picture of the object is no smaller than the variance for subjects who did see a picture. The alternative hypothesis is that the variance for subjects who saw a picture is smaller than the variance for subjects who did not see a picture. Use a level of significance of 0.05.

Short answer

Reject the null in favor of the alternative at level 0.05

Step-by-step solution

Given the information

Let X be the time taken to fuse a random dot stereogram for the first group.

Let Y be the time taken to fuse a random dot stereogram for the second group.
The sample size for the first group is m=43 and for the second group, it is n=35.

Finding the null hypothesis by using the level of significance of 0.5.

The hypotheses under scrutiny are

H₀ : σ₁² ≤ σ₂² H₁ : σ₁² >σ₂²

Then, under the observed data, From Exercise 11 S_x²= 2745.7 and S_y²=783.9.

The F statistic is

\begin{aligned}F=\frac{_S{x}^{2}}{m}\times\frac{n}{s{_{y}}^{2}}\end{aligned}

\begin{aligned}=\frac{2745.7}{43}\times\frac{35}{783.9}\end{aligned}

\begin{aligned}=\frac{63.853}{22.397}\end{aligned}

\begin{aligned}=2.850\end{aligned}

F=2.850

Test statistics follows a null F distribution with degrees of freedom 42 and 34 respectively. Reject the null hypothesis for large values of F, or precisely, if

F< F_42,34^-1(0.95).

Here, F=2.850

While the cutoff value isF< F_42,34^-1(0.95)= 1.737, By using F table.

Since the F statistic exceeds the cutoff value, reject the null at level 0.05 in favor of the alternative.