Question

a sociologist develops a test to measure attitudes towards public transportation and 27 randomly selected subjects are given the test.their mean score is 76.2 and their standard deviation is 21.4. construct the 95% confidence interval for the mean score of all such subjects

Answer 1

Answer:

The 95% confidence interval for the mean score is:

$\mu$ ∈ $(67.73,\ \ 84.67)$

Step-by-step explanation:

We want to estimate the population mean, that is, the average score of all the subjects.

An estimator for the population mean μ is ${\displaystyle {\overline {x}}}$

Since the sample size is $27$ then we use the  statistic  $t_{\frac{\alpha}{2}}$  with $27-1 = 26$ degrees of freedom

The confidence interval for the mean is:

$({\displaystyle {\overline {x}}} + t_{\frac{\alpha}{2}}\frac{S}{\sqrt{n}},\ \ {\displaystyle {\overline {x}}} - t_{\frac{\alpha}{2}}\frac{S}{\sqrt{n}})$

Where

${\displaystyle {\overline {x}}}=76.2$ is the average score of the sample

$S=21.4$ is the standard deviation of the sample

$n= 27$ is the sample size

$t_{\frac{\alpha}{2}} = 2.056$ for a confidence level of 95% and 26 degrees of freedom

Then confidence interval is

$(76.2 + 2.056\frac{21.4}{\sqrt{27}},\ \ {76.2 - 2.056\frac{21.4}{\sqrt{27}})$

$(76.2 + 8.467,\ \ 76.2 - 8.467)$

$\mu$ ∈ $(67.73,\ \ 84.67)$