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, female subjects were asked to rate the attractiveness of their male dates, and a sample of the results is listed below (1 = not attractive; 10 = extremely attractive). Use a 99% confidence level. Can the result be used to estimate the mean amount of attractiveness of the population of all adult males?

5 8 3 8 6 10 3 7 9 8 5 5 6 8 8 7 3 5 5 6 8 7 8 8 8 7

Short answer

The 99% confidence interval for attractiveness is between 5.5 and 7.6.

The data deviated from the normal pattern along with an outlier, and hence, the interval must not be used for concluding results. Also, the data is observed for male dates and may not be valid for the entire adult male population.

Step-by-step solution

Given information

The sample of 26 ratings is observed such that each rating varies from 1 to 10. The confidence level is 99%.

Check the requirements

The necessary conditions 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 26, which is less than 30, the condition for normality will only be satisfied if the data follows a normal distribution.

Assume that the requirements hold true.

Compute the degree of freedom and the critical value

The degree of freedom is computed as follows.

Substitute 26 for in the above formula and simplify.

$$\begin{gathered} df=n-1 \\ =26-1 \\ =25 \end{gathered}$$

The level of significance is 0.01 for the confidence level of 0.99.

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

Use the t-distribution table to obtain the critical value when $\alpha =0.01$and $df=25$.

The value corresponding to row 25 and column 0.01 (two-tailed) is obtained as $t_{\frac{0.01}{2}}=2.787$.

Compute the margin of error 

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

The sample mean can be obtained using the formula $\overline{x}=\frac{1}{26}\sum _{i=1}^{26}{x}_i$, where$x_i$represents the data points in a sample.

Compute the sample mean.

$$\begin{gathered} \overline{x}=\frac{\left(5+8+3...+7\right)}{26} \\ =\frac{171}{26} \\ =6.5769 \end{gathered}$$

Calculate the sample variance using the formula $s^2=\frac{1}{26-1}\sum _{i=1}^{26}{\left(x_i-\overline{x}\right)}^2$.

x	${\left(x-\overline{x}\right)}^2$
5	2.56
8	1.96
3	12.96
8	1.96
6	0.36
10	11.56
3	12.96
7	0.16
9	5.76
8	1.96
5	2.56
5	2.56
6	0.36
8	1.96
8	1.96
7	0.16
3	12.96
5	2.56
5	2.56
6	0.16
8	1.96
7	0.16
8	1.96
8	1.96
8	1.96
7	0.16
	$\sum _{i=1}^{26}{\left(x_i-\overline{x}\right)}^2=88.16$

Subsitute $\sum _{i=1}^{26}{\left(x_i-\overline{x}\right)}^2=88.16$in the formula $s^2=\frac{1}{25-1}\sum _{i=1}^{26}{\left(x_i-\overline{x}\right)}^2$. So,

$$\begin{gathered} s^2=\frac{1}{25}\times 88.16 \\ =\frac{88.16}{25} \\ =3.526 \end{gathered}$$

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

$$\begin{gathered} s=\sqrt{3.526} \\ =1.8798 \end{gathered}$$

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,

$$\begin{gathered} E=2.787\times \frac{1.8798}{\sqrt{26}} \\ =1.0276 \end{gathered}$$

Construct the confidence interval 

The 99% confidence interval is given as follows.

$$\begin{gathered} \overline{x}-E<\mu <\overline{x}-E \\ 6.5769-1.0276<\mu <6.5769+1.0276 \\ 5.5493<\mu <7.6046 \end{gathered}$$

Thus, the 99% confidence level for the mean attractiveness of males is between 5.5 and 7.6.

The mean attractiveness of male dates is based on the ratings of female subjects, which cannot be used to infer about the population of all adult males.

Check the normality of the sample data   

Construct the normal probability plot to check if the sample data follows a normal distributionto check the applicability of the confidence interval.

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

Thus, there exists an outlier, and the values do not follow a straight-line pattern. Thus, it can be inferred that the confidence interval is not useful for estimating the mean.