Question

Confidence Intervals. In Exercises 9–24, construct the confidence interval estimate of the mean.

In a study of speed dating conducted at Columbia University, male subjects were asked to rate the attractiveness of their female dates, and a sample of the results is listed below (1 = not attractive; 10 = extremely attractive). Use a 99% confidence level. What do the results tell us about the mean attractiveness ratings made by the population of all adult females?

7 8 2 10 6 5 7 8 8 9 5 9

Short answer

The 99% confidence interval for the mean attractiveness is between and .

The confidence interval determines the attractiveness of the females that the male subjects dated. Thus, it does not tell anything about the population of adult females.

Step-by-step solution

Given information

The sample of 12 ratings is observed from male subjects towards the attractiveness of their female dates. The confidence level is 99%.

Check the requirements

The conditions necessary for using any sample data to construct confidence intervals are as follows.

The sample is collected from the population of females that satisfies the condition of a simple random sampling.As the sample size is 12, which is less than 30, the condition for normality will only be satisfied if the data follows a normal distribution. This can be verified from the normal probability plot.

The steps to construct the normality plot are as follows.

1. Draw two axes for the observed values and z-scores.
2. Mark the observations corresponding to the z-scores.
3. Observe if the marks follow a straight-line pattern.

Thus, the assumptions are satisfied.

Compute the degree of freedom and the critical value

The degree of freedom is computed as follows.

$$\begin{gathered} df=n-1 \\ =12-1 \\ =11 \end{gathered}$$

The level of significance corresponding to the 99% confidence level is obtained as

$$\begin{gathered} \alpha =1-0.99 \\ =0.01 \end{gathered}$$

Use the t-distribution table to obtain the critical value at $\alpha =0.01$and $df=11$.

The value is obtained at the intersection of row 11 and column 0.01 (two-tailed area), which is$t_{\frac{0.01}{2}}=3.106$.

Compute the margin of error 

Let $X$be the random variable that denotes the rating of females.

Compute the sample mean.

$$\begin{gathered} \overline{x}=\frac{\sum x_i}{n} \\ =\frac{7+8+2+...+9}{12} \\ =\frac{84}{12} \\ =7 \end{gathered}$$

Calculate the sample variance using the formula $s^2=\frac{1}{n-1}\sum _{i=1}^{12}{\left(x_i-\overline{x}\right)}^2$.

$x$	${\left(x-\overline{x}\right)}^2$
7	0
8	1
2	25
10	9
6	1
5	4
7	0
8	1
8	1
9	4
5	4
9	4
	$\sum _{i=1}^{12}{\left(x_i-\overline{x}\right)}^2=54$

Subsitute $\sum _{i=1}^{12}{\left(x_i-\overline{x}\right)}^2=54$in the formula $s^2=\frac{1}{12-1}\sum _{i=1}^{12}{\left(x_i-\overline{x}\right)}^2$. So,

.

$$\begin{gathered} s^2=\frac{1}{11}\times 54 \\ =\frac{54}{11} \end{gathered}$$

The square root of the sample variance is equal to the sample standard deviation. Thus, the sample standard deviation is given as follows.

$$\begin{gathered} s=\sqrt{\frac{54}{11}} \\ =2.2156 \end{gathered}$$

The margin of error is given by the formula $E=t_{\frac{\alpha }{2}}\times \frac{s}{\sqrt{n}}$.Substitute the respective value obtained from above in the equation and simplify to compute the margin of error. So,

$$\begin{gathered} E=3.106\times \frac{2.2156}{\sqrt{12}} \\ =1.9865 \end{gathered}$$

Construct the confidence interval 

The confidence interval is given as follows.

$$\begin{gathered} \overline{x}-E<\mu <\overline{x}-E \\ 7-1.9865<\mu <7+1.9865 \\ 5.0135<\mu <8.9865 \\ 5.0<\mu <9.0 \end{gathered}$$

The 99% confidence interval for the mean attractiveness ratings is between 5.0 and 9.0.

Analyze the confidence interval   

Therefore, the 99% confidence interval for the mean attractiveness of females dated by the male subjects is between 5.0 and 9.0. The ratings do not state the attractiveness of all adult females, as it is specific to the women dated by certain males.