Question

A diet doctor claims Australians are, on average, overweight by more than 10kg. To test this claim, a random sample of 100 Australians were weighed, and the difference between their actual weight and their ideal weight was calculated and recorded.

The data are contained in the Excel file Weights.xlsx.

Use these data to test the doctor's claim at the 5% level of significance.

Excess weight (Kgs)
16.0
4.0
4.0
4.5
11.0
7.0
7.0
16.5
14.5
5.5
16.5
0.5
13.5
26.0
28.0
31.5
14.0
25.0
14.5
1.0
2.5
4.0
17.5
6.0
5.0
4.5
10.0
11.0
8.0
0.5
4.5
10.5
31.0
23.0
11.5
10.0
10.0
22.5
4.0
12.5
29.5
23.5
10.5
10.5
10.0
12.5
21.5
5.0
5.0
20.0
15.0
15.0
25.0
15.0
11.0
28.5
14.0
24.5
20.0
7.5
1.5
5.5
9.5
3.0
8.5
4.0
5.5
8.5
17.0
13.0
20.5
23.0
18.5
16.5
6.5
5.0
16.5
5.0
9.0
15.0
21.0
9.0
24.0
8.0
9.0
6.5
23.0
7.5
14.5
15.5
0.5
10.0
23.0
21.0
7.5
15.0
10.5
8.5
16.5
17.0

Question 10

(Part B)

In this question, we let μ represent

	a.	the population mean 12.7
	b.	the population average ideal weight of Australians
	c.	the population average actual weight of Australians
	d.	the population average of difference between the actual and ideal weights
	e.	None of the above

Question 11

(Part B)

The null hypothesis is

	a.	H0: μ > 10
	b.	H0: μ = 10
	c.	H0: μ = 12.7
	d.	H0: μ < 12.7
	e.	None of the above

Question 12

(Part B)

The alternative hypothesis is

	a.	HA: μ > 10
	b.	HA: μ < 12.7
	c.	HA: μ ≠ 10
	d.	HA: μ ≠ 12.7
	e.	None of the above

Question 13

(Part B)

The value of the t-statistic is

	a.	–3.527
	b.	0.3527
	c.	3.527
	d.	–0.275
	e.	None of the above

Question 14

(Part B)

The decision rule is

	a.	reject HA if t > 1.984
	b.	reject H0 if t > 1.984
	c.	reject H0 if t < 1.660
	d.	reject H0 if t > 1.660
	e.	None of the above

Question 15

(Part B)

The p-value is

	a.	1.660
	b.	0.05
	c.	0.0003
	d.	0.0070
	e.	None of the above

Answer 1

	Values ( X )	Σ ( Xi- X̅ )2
	16	10.7912
	4	75.9512
	4	75.9512
	4.5	67.4862
	11	2.9412
	7	32.6612
	7	32.6612
	16.5	14.3262
	14.5	3.1862
	5.5	52.0562
	16.5	14.3262
	0.5	149.2062
	13.5	0.6162
	26	176.4912
	28	233.6312
	31.5	352.8762
	14	1.6512
	25	150.9212
	14.5	3.1862
	1	137.2412
	2.5	104.3462
	4	75.9512
	17.5	22.8962
	6	45.0912
	5	59.5212
	4.5	67.4862
	10	7.3712
	11	2.9412
	8	22.2312
	0.5	149.2062
	4.5	67.4862
	10.5	4.9062
	31	334.3412
	23	105.7812
	11.5	1.4762
	10	7.3712
	10	7.3712
	22.5	95.7462
	4	75.9512
	12.5	0.0462
	29.5	281.7362
	23.5	116.3162
	10.5	4.9062
	10.5	4.9062
	10	7.3712
	12.5	0.0462
	21.5	77.1762
	5	59.5212
	5	59.5212
	20	53.0712
	15	5.2212
	15	5.2212
	25	150.9212
	15	5.2212
	11	2.9412
	28.5	249.1662
	14	1.6512
	24.5	138.8862
	20	53.0712
	7.5	27.1962
	1.5	125.7762
	5.5	52.0562
	9.5	10.3362
	3	94.3812
	8.5	17.7662
	4	75.9512
	5.5	52.0562
	8.5	17.7662
	17	18.3612
	13	0.0812
	20.5	60.6062
	23	105.7812
	18.5	33.4662
	16.5	14.3262
	6.5	38.6262
	5	59.5212
	16.5	14.3262
	5	59.5212
	9	13.8012
	15	5.2212
	21	68.6412
	9	13.8012
	24	127.3512
	8	22.2312
	9	13.8012
	6.5	38.6262
	23	105.7812
	7.5	27.1962
	14.5	3.1862
	15.5	7.7562
	0.5	149.2062
	10	7.3712
	23	105.7812
	21	68.6412
	7.5	27.1962
	15	5.2212
	10.5	4.9062
	8.5	17.7662
	16.5	14.3262
	17	18.3612
Total	1271.5	5866.625

Mean X̅ = Σ Xi / n
X̅ = 1271.5 / 100 = 12.715

Sample Standard deviation S_X = √ ( (Xi - X̅ )2 / n - 1 )
S_X = √ ( 5866.625 / 100 -1 ) = 7.698

Question 10

a.	the population mean 12.7

Question 11

b.	H0: μ = 10

Question 12

a.	HA: μ > 10

Question 13

Test Statistic :-
t = ( X̅ - µ ) / (S / √(n) )
t = ( 12.715 - 10 ) / ( 7.698 / √(100) )
t = 3.5269

c.

3.527

Question 14

Test Criteria :-
Reject null hypothesis if t > t(α, n-1)
t(α, n-1) = t(0.05 , 100-1) = 1.66
t > t(α, n-1) = 3.5269 > 1.66
Result :- Reject null hypothesis

d.	reject H0 if t > 1.660

Question 15

P - value = P ( t > 3.5269 ) = 0.0003
Looking for the value t = 3.257 in t table across n - 1 = 100 - 1 = 99 degree of freedom.

c.

0.0003