8.54 / 8.75 =0.975
1 - 0.975 = 0.025
0.025 x 100% ( and nearest thenth)
=2.5%
Using the t-distribution, it is found that:
a. The <u>margin of error</u> is of 4.7 homes.
b. The 98% confidence interval for the population mean is (19.3, 28.7).
The information given in the text is:
- Sample mean of
. - Sample standard deviation of
. - Sample size of
.
We are given the <u>standard deviation for the sample</u>, which is why the t-distribution is used to solve this question.
The confidence interval is:
The margin of error is:
Item a:
The critical value, using a t-distribution calculator, for a two-tailed <u>98% confidence interval</u>, with 23 - 1 = <u>22 df</u>, is t = 2.508.
Then, the <em>margin of error</em> is:
Item b:
The interval is:
The 98% confidence interval for the population mean is (19.3, 28.7).
A similar problem is given at brainly.com/question/15180581
Answer:
b. The student's scores on the posttest would have a smaller standard deviation.
Step-by-step explanation:
The first test is taken before the material is covered in class so we expect the standard deviation to be high because not everyone's scores would be lying close to the mean. Equal number of students mastered most, some or almost none of the material from reading the textbook based on the pretest result. this means the data is varying, so the standard deviation is large.
Whereas, after the teacher has taught the material and given the homework, they must have understood most of the material. The test they take after teaching as a post test. The results of the post test would have a smaller standard deviation as most of the students would have scored good. Hence, the student's scores on the posttest would have a smaller standard deviation.
Answer:
1:1.5 1 part lemon to 1.5 parts sugar
1.5 : 7 1.5 parts sugar to 7 parts water
Step-by-step explanation:
