Question

A town council wants to estimate the proportion of residents who are in favor of a proposal to upgrade the computers in the town library. A random sample of 100 residents was selected, and 97 of those selected indicated that they were in favor of the proposal. Is it appropriate to assume that the sampling distribution of the sample proportion is approximately normal?

Answer 1

Answer:

We need to check the conditions in order to use the normal approximation.

1) $np=100*0.97=97 > 10$

2) $n(1-p)=100*(1-0.97)=3 < 10$

For this case the first condition is satisfied but the second NO, so then is not appropiate for this case use the normal approximation given by:

$p \sim N(p , \sqrt{\frac{p(1-p)}{n}})$

Step-by-step explanation:

For this case we have the following info given:

n = 100 represent the sample size

x= 97 represent the people selected who  indicates that they were in favor of the proposal

The estimated proportion for this case is given by:

$\hat p =\frac{x}{n}= \frac{97}{100}=0.97$

We need to check the conditions in order to use the normal approximation.

1) $np=100*0.97=97 > 10$

2) $n(1-p)=100*(1-0.97)=3 < 10$

For this case the first condition is satisfied but the second NO, so then is not appropiate for this case use the normal approximation given by:

$p \sim N(p , \sqrt{\frac{p(1-p)}{n}})$

Answer 2

Answer:

Since n(1-p) < 5, it is not appropriate to assume that the sampling distribution of the sample proportion is approximately normal.

Step-by-step explanation:

Binomial probability distribution:

Probability of x sucesses on n repeated trials, with p probability.

Can be approximated to the normal distribution if:

np > 5 and n(1-p) > 5

In this problem:

A random sample of 100 residents was selected, and 97 of those selected indicated that they were in favor of the proposal.

This means that $n = 100, p = \frac{97}{100} = 0.97$

Is it appropriate to assume that the sampling distribution of the sample proportion is approximately normal?

np = 100*0.97 = 97 > 5

n(1-p) = 100*0.03 = 3 < 5

Since n(1-p) < 5, it is not appropriate to assume that the sampling distribution of the sample proportion is approximately normal.