Question

3. In a March 2010 poll of 1600 American adults, 1088 described themselves as "totally happy" (this category included both “very happy" or "somewhat happy” responses) Use STATCRUNCH to construct a 95% confidence interval for the proportion of American adults who describe themselves as “totally happy." a. Best Point Estimate b. Margin of Error for a 95% confidence level. c. Construct the 95% confidence interval for the proportion of American adults who describe themselves as “totally happy”. Statement: (Round to 4 decimal places.) d. Can we safely say that more than half of American adults describe themselves as "totally happy”?

Answer 1

Answer)

A)

Best point estimate P = 1088/1600 = 0.68

B)

N = 1088

P = 0.68

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 1088

N*(1-p) = 512

Both the conditions are met so we can use standard normal z table to estimate the interval

Critical value z from z table for 95% confidence level is 1.96

Margin of error (MOE) = Z*√P*(1-P)/√N = 0.02285733142

C)

Confidence interval is given by

P-MOE < P < P+MOE

0.65714266857 < P < 0.70285733142

We are 95% confident that true population proportion lies in between 0.6571 and 0.7029

D)

Null hypothesis Ho : P = 0.5

Alternate hypothesis Ha : P > 0.5

As the null hypothesised value is not in the interval

We can reject the null hypothesis Ho

So we have enough evidence to conclude that more than half of Americans adults describe themselves as totally happy