Question

A politician want to know his constituents' attitudes toward banning plastic bags. To test this he asks 120 constituents their attitudes toward the potential ban on a 5-point scale (1 = strongly oppose, 2 = oppose, 3 = neither oppose nor support, 4 = support, 5 = strongly support) and finds the following: M = 4.22, SD = .89 What are his constituents' attitudes toward banning plastic bags α = .01? Round to the second decimal place (every calculation). Complete all reporting results steps.

What is the est SEM?

What is t?

Do you reject or fail to reject the null?

Explain the results in non-statistical language.

Answer 1

Null Hypothesis H0: The constituents' attitudes is neutral towards banning plastic bags. Mean Score = 3

Alternative Hypothesis Ha: The constituents' attitudes strongly support towards banning plastic bags. Mean Score > 3

SEM = s / = 0.89 / = 0.08

t = (4.22 - 3) / 0.08 = 15.25

Degree of freedom = n-1 = 120-1 = 119

Critical value of t at α = .01 and df = 119 is 2.62

Since the observed F (15.25) is less than the critical value, we reject the null hypothesis H0.

There is significant evidence to conclude that constituents' attitudes strongly support towards banning plastic bags.