Question

19) Five people who were convicted of speeding were ordered by the court to attend a workshop. A special device put into thei
Name: RT 70.00 66.00 -4 JM 68.00 62.00 -6 Step 1: Research hypothesis (2pt): thir who attend the workshop will have lower dri

I need help with step 2. I dont know how to do the table!
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Answer #1

19)

Suppose, random variable X and Y denote maximum speed before and after workshop respectively. Also random variable D denotes difference in maximum speeds as D=X-Y.

We are given data in pairs corresponding to each sample unit and so we have to perform paired t-test.

We have to test for null hypothesis H_0:\bar d=0

against the alternative hypothesis H_1:\bar d>0

Participants Before (x) After (y) d=x-y d-M = d-24/5 = d-4.8 (d-M)2
LB 65 60 5 0.2 0.04
JK 70 65 5 0.2 0.04
RC 60 56 4 -0.8 0.64
RT 70 66 4 -0.8 0.64
JM 68 62 6 1.2 1.44
Total - - 24 - 2.80

Here, number of observation n=5

Mean was calculated as

M =d = id : 4 = 4.8 d) = 4.8

Sample standard deviation of difference (d) is given by

s_d=\frac{1}{n-1}\sum_{i=1}^{n}(d_i-\bar d)^2=\frac{1}{4}\sum_{i=1}^{5}(d_i-\bar d)^2=\frac{2.8}{4}=0.7

Our test statistic is given by

t=\frac{\bar d}{s_d/\sqrt n}

\therefore t_{calculated}=\frac{4.8}{0.7/\sqrt 5}=15.33304

Level of significance \alpha = 0.05

Degrees of freedom v=n-1=5-1=4

In case of right tailed (greater than type alternative hypothesis) test critical value is given by

t_{critical}=t_{\alpha, v}=t_{0.05,4}=2.131847 [As P(t4>2.131847) = 0.05]

In this type of (right tailed) test we reject null hypothesis if t_{calculated}>t_{critical} .

Here we observe that t_{calculated}=15.33304>2.131847=t_{critical} .

So, we reject our null hypothesis.

Hence, based on the given data we conclude that people are likely to drive more slowly after this type of workshop.

A key note- In the given images of question and attempted answer, I noticed that in Research hypothesis it is mentioned that lower driving speed than those who don't attend the workshop. However, it must not be so. There is no comparison with other people. the comparison is between maximum speed before and after workshop for each individual attending the workshop (as clearly mentioned in the given problem statement). Thus mistake was done in constructing null hypothesis also.

However, I would like to rectify and restate Research and Null hypothesis (in terms of language) as follows.

Null hypothesis- Those who attended the workshop has no change in driving speed between before and after attending the workshop.

Alternative hypothesis- Those who attended the workshop has lower driving speed after attending the workshop than before the workshop.

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