# A simple confidence problem that I can't seem to solve

#### kulmser

##### New Member
Mr. Grumpy and Mr. Happy are both running for Governor. Mr. Grumpy will eventually win the election with 51 percent of the vote. A day before the election, a state-wide newspaper surveys 100 people about their choice for governor. Assume that survey respondents accurately report who they will eventually vote for. What is the probability Mr. Happy will be supported by 51 percent or more of the survey respondents.

Now I can see that Mr Grumpy will win with 51% of the vote. xi=1 for people who prefer Mr. Happy. and x(bar) is the sample mean of xi. True expected value of the survey response is 0.49.

xbar is the survey response = 0.51
u is the true sample mean (0.49)
σ/n.5 is the sample standard error of xbar. However I don't seem to be able to figure out the σ. How could I get the σ??

#### GretaGarbo

##### Human
Look at the binomial distribution. What is the variance of the sum of voter for Mr Grumpy? And what is the variance for the proportion?

#### hlsmith

##### Less is more. Stay pure. Stay poor.
I agree with Greta on this being a binomial problem, that and you have to assume this is a random sample for conclusion purposes.

#### kulmser

##### New Member
Look at the binomial distribution. What is the variance of the sum of voter for Mr Grumpy? And what is the variance for the proportion?
Since it is np(1-p)
Shouldnt it be

Sqrt(np(1-p))/(n^0.5))?

However the answer I got from the lecturer was as following.
Sample standard error of xbar = [(0.51*0.49)/n]^0.5
How could this be?

#### GretaGarbo

##### Human
What would p be in your example?

Do you know that sqrt(x) = x^0.5 ?