Question

A hotel manager believes that 27% of the hotel rooms are booked. If the manager is correct, what is the probability that the proportion of rooms booked in a sample of 423 rooms would differ from the population proportion by less than 6%

Answer 1

The probability that the proportion of rooms booked in a sample of 423 rooms would differ from the population proportion by less than 6% is 0.9946.

Explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

`\mu_(\hat p)=p`

The standard deviation of this sampling distribution of sample proportion is:

`\sigma_(\hat p)=\sqrt{(\hat p(1-\hat p))/(n)}`

The information provided here is:

p = 0.27

n = 423

As n = 423 > 30, the sampling distribution of sample proportion can be approximated by the Normal distribution.

The mean and standard deviation of the sampling distribution of sample proportion are:

`\mu_(\hat p)=p=0.27\\\\\sigma_(\hat p)=\sqrt{(\hat p(1-\hat p))/(n)}=\sqrt{(0.27*(1-0.27))/(423)}=0.0216`

Compute the probability that the proportion of rooms booked in a sample of 423 rooms would differ from the population proportion by less than 6% as follows:

`P(|\hat p-p|<0.06)=P(p-0.06<\hat p<p+0.06)`

`=P(0.27-0.06<\hat p<0.27+0.06)\\\\=P(0.21<\hat p<0.33)\\\\=P((0.21-0.27)/(0.0216)<(\hat p-\mu_(\hat p))/(\sigma_(\hat p))<(0.33-0.27)/(0.0216))\\\\=P(-2.78<Z<2.78)\\\\=P(Z<2.78)-P(Z<-2.78)\\\\=0.99728-0.00272\\\\=0.99456\\\\\approx 0.9946`

*Use a z-table.

Thus, the probability that the proportion of rooms booked in a sample of 423 rooms would differ from the population proportion by less than 6% is 0.9946.