Question

Dentists’ use of laughing gas. According to the American Dental Association, 60% of all dentists use nitrous oxide (laughing gas) in their practice. In a random sample of 75 dentists, let $\hat{p}$represent the proportion who use laughing gas in practice.

a. Find $E\left(\hat{p}\right)$.

b. Find ${\sigma }_{\hat{p}}$.

c. Describe the shape of the sampling distribution of $\hat{p}$.

d. Find $P\left(\hat{p}>0.70\right)$.

Short answer

a. The mean of the sample proportion is $E\left(\hat{p}\right)=0.60$.

b. The standard deviation of the sample proportion is 0.6557.

c. The shape of the sampling distribution of the sample proportion is approximately symmetric (normal).

d. The required probability is 0.4404.

Step-by-step solution

Given information

The proportion of all dentists that uses nitrous oxide in their practice is p = 0.60.

A random sample of size n = 75 is selected.

Let $\hat{p}$represents the sample proportion that uses laughing gas in practice.

Computing the mean of the sample proportion

a

The mean of the sample proportion is obtained as:

$$\begin{gathered} E\left(\hat{p}\right)=p \\ =0.60 \end{gathered}$$.

Since $E\left(\hat{p}\right)=p$.

Therefore, $E\left(\hat{p}\right)=0.60$.

Computing the standard deviation of the sample proportion

b.

The standard deviation of the sample proportion is obtained as:

$$\begin{gathered} {\sigma }_{\hat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}} \\ =\sqrt{\frac{0.60\times 0.40}{75}} \\ =\sqrt{\frac{32.24}{75}} \\ =\sqrt{0.4299} \\ =0.6557 \end{gathered}$$.

Therefore,${\sigma }_{\hat{p}}=0.6557$.

Determining the shape of the sampling distribution

c.

Here,

$$\begin{gathered} n\hat{p}=75\times 0.60 \\ =45 \\ >15 \end{gathered}$$,

and

$$\begin{gathered} n\left(1-\hat{p}\right)=75\times 0.40 \\ =30 \\ >15 \end{gathered}$$

The conditions to use normal approximation are satisfied. Therefore, the shape of the sampling distribution of the sample proportion is approximately symmetric (normal).

Computing the required probability

d.

To find the required probability normal approximation is used.

Therefore,

$$\begin{gathered} P\left(\hat{p}>0.70\right)=P\left(\frac{\hat{p}-p}{{\sigma }_{\hat{p}}}>\frac{0.70-p}{{\sigma }_{\hat{p}}}\right) \\ =P\left(Z>\frac{0.70-0.60}{0.6557}\right) \\ =P\left(Z>\frac{0.10}{0.6557}\right) \\ =P\left(Z>0.15\right) \\ =1-P\left(Z⩽0.15\right) \\ =1-0.5596 \\ =0.4404 \end{gathered}$$

The z-table is used to find the probability; the value at the intersection of 0.10 and 0.05 is the probability of a z-score less than or equal to 0.15.

Hence, the required probability is 0.4404.