Question

7. Suppose the National Transportation Safety Board (NTSB) wants to examine the safety of compact cars, midsize cars, and full-size cars. It collects a sample of three for each of the treatments (cars types). The hypothetical data provided below from 10 trials report the mean pressure applied to the driver’s head during a crash test for each type of car.

Compact: 635, 671, 648, 685, 648, 651, 654, 682, 687, 627

Midsize: 482, 529, 541, 518, 497, 526, 507, 492, 499, 451

Full-size: 451, 483, 464, 447, 456, 499, 484, 492, 449, 449

10. An instructor teaching algebra 1 to ninth-grade students wishes to analyze the difference between student achievement before and after the implementation of an online help resource. For 6 weeks, students worked with conventional, in-class and homework resources, and then for the next 6 weeks, an online help desk was made available to them. The scores for 6 students on a district benchmark test before and after the implementation of the online help resource are listed below.

Before: 22, 18, 33, 20, 23, 27

After: 28, 21, 32, 25, 33, 28



12. A college counselor wonders whether second semester students take fewer units than first semester students. From the population of each group (first semester and second semester), she selects 10 students at random. The following data were collected:

First semester students: 10, 12, 14, 14, 15, 15, 15, 16, 16, 18

Second semester students: 6, 9, 9, 10, 12, 12, 13, 14

Answer 1

7. for the given data: the respective means can be calculated as follows:

-> Compact: 635, 671, 648, 685, 648, 651, 654, 682, 687, 627

mean= (635+671+648+685+648+651+654+682+687+627)/10

=6588/10

=658.8

Midsize: 482, 529, 541, 518, 497, 526, 507, 492, 499, 451

mean= (482+529+541+518+497+526+507+492+499+451)/10

=5041/10

=504.1

Full-size: 451, 483, 464, 447, 456, 499, 484, 492, 449, 449

mean= (451+483+464+447+456+499+484+492+449+449)/10

=4674/10

=467.4

10. for the given data the mean before and after the intervention is as follows:

Before: 22, 18, 33, 20, 23, 27

mean1= (22+18+33+20+23+27)/6

=143/6

=23.83

After: 28, 21, 32, 25, 33, 28

mean2= (28+21+32+25+33+28)/6

=167/6

=27.83

The null hypothesis (H0) assumes that the true mean difference (μd) is equal to zero.

H0: μd = 0

using paired t test for difference of mean in spss we found that p value is 0.054 which is greater than 0.05 hence we do not reject the null hypothesis hnce, the means are almost same or there is no difference.

12. for the given data

First semester students: 10, 12, 14, 14, 15, 15, 15, 16, 16, 18

Second semester students: 6, 9, 9, 10, 12, 12, 13, 14

again using spss and unpaired two sample t test we get:

now the significance level was less than 0.05 hence we reject the null hypothesis.