Question

1. Researchers conducted a study to determine that having a gene disruption may be related to an increased risk of developing dyslexia. The two-way table indicates whether those having the gene disruption have a different proportion between having dyslexia versus no dyslexia.

	Dyslexia	No Dyslexia (control)
Gene disruption	18	15
No gene disruption	99	185
Total	117	200

1. Calculate the statistic (with the notation) you will need to perform a hypothesis test.

2. Perform a hypothesis test to determine whether there is a difference in proportion of those with the gene disruption who have dyslexia versus those with no dyslexia

1. State the null and alternative hypotheses. DRAW/LABEL/SHADE a curve, and identify the resulting p-value.

2. At a =.05 significance level, what is your conclusion for your test in context of the problem?

Answer 1

a) the sample proportions with gene disruption is computed here as:

p₁ = 18 / 117 = 0.1538
p₂ = 15/ 200 = 0.075

The pooled proportion is computed here as:
P = (x₁ + x₂) / (n₁ + n₂) = (18 + 15) / (117 + 200) = 0.1041

The standard error here is computed as:

The test statistic here is computed as:

Therefore 2.2169 is the test statistic value here.

b) As we are testing here whether the two proportions are different, therefore this is a two tailed test, the null and the alternative hypothesis here are given as:

The p-value here is computed from the standard normal tables as:
p = 2P(Z > 2.2169) = 2*0.0133 = 0.0266

Therefore 0.0266 is the required p-value here.

This is given on a plot as:

As the p-value here is 0.0266 < 0.05 which is the level of significance, therefore the test is significant here and we can reject the null hypothesis here and conclude that we have sufficient evidence of difference between the two proportions.