Question

A random sample of 121 NKU students was taken to test the proportion of students who had attended a Reds game in 2015.

95% confidence interval results:

Variable	Count	Total	Sample Prop.	Std. Err.	L. Limit	U. Limit
Reds	65	121	0.53719008	0.045328634	0.44834759	0.62603257

1. Verify that the conditions are met to construct a 95% confidence interval for the proportion of NKU students who attended a Reds game in 2015.

2. Construct a 95% confidence interval:

3. Interpret the above interval:

4. Can one conclude that a majority of NKU students attended a Reds game in 2015?

Answer 1

1)

Answer)

N = 121

P = 65/121

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 = 65

N*(1-p) = 56

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

2)

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

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

Interval is given by

(P-MOE, P+MOE)

(0.44834759, 0.62603257)

3)

We are 95% confident that true population proportion lies in between 0.45 and 0.63

4)

Null hypothesis Ho : P = 0.5

Alternate hypothesis Ha : p > 0.5

As the interval contains the null hypothesised value 0.5

We fail to reject the null hypothesis Ho

So we do not have enough evidence to conclude that majority of NKU students attended a Reds game in 2015