A sociologist develops a test to measure attitudes about 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 90% confidence interval for the mean score of all such subjects. Immersive Reader

The 90% confidence interval for the mean score of all such subjects is between 39.7 and 112.7

Step-by-step explanation:

We have the standard deviation of the sample, so we use the t-distribution to build the confidence interval.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 27 - 1 = 26

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 26 degrees of freedom(y-axis) and a confidence level of ). So we have T = 1.7056

The margin of error is:

M = T*s = 1.7056*21.4 = 36.50.

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 76.2 - 36.5 = 39.7

The upper end of the interval is the sample mean added to M. So it is 76.2 + 36.5 = 112.7

The 90% confidence interval for the mean score of all such subjects is between 39.7 and 112.7

the inequality is:

250c+180g≥950

step-by-step explanation:

let c denotes the number of days in a week   miyoko worked as a cryptographer.

and g denotes the number of days in a week she works as a geologist.

now we are given that she earns $250 for every day she works as a cryptographer,and earns $180 for every day she works as a geologist.

now miyoko's goal is to earn more than $950 this week.

hence, the inequality could be represented as:

250c+180g≥950