Two random samples are taken one from among firstyear UVA st

Two random samples are taken, one from among first-year UVA students and the other from among fourth-year UVA students. Both samples are asked if they favor modifying the Honor Code. A summary of the sample sizes and number of each group answering yes\'\' are given below: First-Years (Pop. 1):Fourth-Years (Pop. 2):n1=88,n2=91,x1=44x2=65 Is there evidence, at an =0.095 level of significance, to conclude that there is a difference in proportions between first-years and fourth-years? Carry out an appropriate hypothesis test, filling in the information requested. A. The value of the standardized test statistic: B. The p-value is

Solution

H₀: P₁ = P₂ ( no significant difference)
H_a: P₁ ? P₂ (significant difference)

n1=98,n2=98,x1=71x2=70

Pooled sample proportion. Since the null hypothesis states that P₁=P₂, we use a pooled sample proportion (p) to compute the standard error of the sampling distribution.

p0 = (p₁ * n₁ + p₂ * n₂) / (n₁ + n₂)

p0 = (71 + 70/ 98 + 98)

= 141/196

Standard error. Compute the standard error (SE) of the sampling distribution difference between two proportions.

SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }

where p is the pooled sample proportion, n₁ is the size of sample 1, and n₂ is the size of sample 2.

hence SE = Square root of 141/ 196 X 55/196 (2/98)

= 2.029

Test statistic. The test statistic is a z-score (z) defined by the following equation.

z = (p₁ - p₂) / SE

where p₁ is the proportion from sample 1, p₂ is the proportion from sample 2, and SE is the standard error of the sampling distribution.

Test statistic = 1/98 divided by 2.029= 5.029

the Pvalue in Z table shows that the probability is negligible. Hence at a significance of 0.065 % the hypothesis is rejected as p value , significnace level

This shows there is significant difference between the 2 sample population proportion choices.