Question

lab scores before treatment
4,3
12,4
4,5
6,7
7,8
8,9
5,6
10,3
6,7
5,4
6,2
17,9
6,5
7,8
12,5
11,2
5,4
8,7
7,8
8,9
6,7
5,4
4,3
12,4
5,4
5,2
14,9
5,4
6,5
10,4
9,4
6,5
10,4
10,4
9,4
10,7
12,4
8,0
13,9
6,7
5,4
10,4
9,4
10,7
10,3
6,7
11,6
5,6

doing sports

yes

no

no

yes

yes

no

no

yes

yes

yes

no

no

yes

yes

no

no

yes

yes

no

yes

yes

no

no

yes

yes

yes

no

no

no

no

yes

no

yes

no

no

yes

yes

no

no

yes

yes

no

yes

yes

no

no

yes

no

1-) the solving method is First step, write your hypothesis carefully, and in full sentences
H 0 : .............
H 1 : ...............
• Second step, conduct your test
• Third step, compare your result with table
• Fourth step, write the result of your test, discuss the result
• Fifth step, indicate type-1 and type-2 errors in full sentences (For the first question,
indicate them for item (a) but not item (b); For the second question, indicate them)

a-) You suspect that people who are doing sports have lab scores lower than 5.0(before the treatment) Conduct a hypothesis test to see if the lab score is significantly lower than 5.0

b-)Conduct a hypothesis test to see if people who are doing sports have lab scores equal to 5.0

c-) Build a 95% confidence interval for lab scores of people who are doing sports and show that it is harmony with (b)

Answer 1

a). Conduct a hypothesis test to see if the lab score for those are doing sports is significantly lower than 5.0.

1.

2.We extract the data corresponding to those who are doing sports and calculate the summary: We assume that the data is from a Normal population and the observations are drawn independently of each other.

	Lab	Sports
	4.3	yes
	6.7	yes
	7.8	yes
	10.3	yes
	6.7	yes
	5.4	yes
	6.5	yes
	7.8	yes
	5.4	yes
	8.7	yes
	8.9	yes
	6.7	yes
	12.4	yes
	5.4	yes
	5.2	yes
	9.4	yes
	10.4	yes
	10.7	yes
	12.4	yes
	6.7	yes
	5.4	yes
	9.4	yes
	10.7	yes
	11.6	yes
Mean	8.120833
SD	2.465762

We have the sample number n=24.

Test statistic: will follow a t-distribution with n-1 df.

Level of significance:

3. The Critical value or the table value of t is -1.714 at 5% and 23 df. Since the calculated value of t>the tabulated value(left tail), we do not reject the null hypothesis.

4. It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean is less than 5, at the 0.05 significance level.

5. For the test above, the p-value or the Type I error is 1>0.05.

b).

1.

.

2.Test statistic:

st statistic: will follow a t-distribution with n-1 df.

Level of significance:

3. The Critical value is tc=2.069. The calculated value of t >the tabulated value.

4. It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean is different than 5, at the 0.05 significance level.

5. The type I error Probability is p=0<0.05.

c. The 95% confidence interval is given by

We can observe that the interval does not include 0, and is bent on the positive side with the lower limit more than 5, indicating that the difference is significant. This is well in line with the decision rule in b.