38.0k views
3 votes
In an opinion poll, 25% of a random sample of 200 people said that they were strongly opposed to having a state lottery. The standard error of the sample proportion is approximately_________.

2 Answers

1 vote

Final answer:

The standard error of the sample proportion is approximately 0.0306.

Step-by-step explanation:

The standard error of a sample proportion can be calculated using the formula:

Standard Error = √((p)(1-p)/n)

where p is the proportion of the sample and n is the sample size. In this case, the proportion is 0.25 (since 25% of the sample is strongly opposed to having a state lottery) and the sample size is 200. Plugging these values into the formula:

Standard Error = √((0.25)(1-0.25)/200) = √(0.1875/200) = √(0.0009375) = 0.0306

So, the standard error of the sample proportion is approximately 0.0306.

User Amittn
by
4.1k points
2 votes

Answer: 0.031 .

Step-by-step explanation:

The standard error of the sample proportion is given by :-


SE_p=\sqrt{(p(1-p))/(n)}

, where p= Sample proportion and n is the sample size.

As per given , we have

In an opinion poll, 25% of a random sample of 200 people said that they were strongly opposed to having a state lottery.

i.e. p= 0.25 and n= 200

Then , the standard error of the sample proportion
=\sqrt{(0.25(1-0.25))/(200)}


=\sqrt{(0.25*0.75)/(200)}=√(0.0009375)\\\\=0.0306186217848\approx0.031

Hence, the standard error of the sample proportion is approximately 0.031 .

User Ruehri
by
3.9k points