Question

he following data values represent the weight changes of teenage girls who are suffering from anorexia, collected from a sample. 1. Estimate the mean weight change of teenage girls who are suffering from anorexia using a point estimate. 2. Construct 99% confidence interval for mean weight change of teenage girls who are suffering from anorexia. 3. Suppose someone claims that the mean weight change of teenage girls who are suffering from anorexia is more than or equal 7.3. Test this claim using an appropriate hypothesis test at 10% significance level. column1 1.7 0.7 -0.1 -0.7 -3.5 14.9 3.5 17.1 -7.6 1.6 column2 11.7 6.1 1.1 -4.0 20.9 -9.3 2.1 1.4 -0.3 -3.7 column3 -1.4 -0.8 2.4 12.6 1.9 3.9 0.1 15.4 -0.7

Answer 1

(a) estimate of mean=sum(x)/n=87/29=3

(b) since the population standard deviation is not known/given, so we use t-value to find the confidence interval

(1-alpha)*100% confidence interval for population mean=sample mean±t(alpha/2,n-1)*sd/sqrt(n)

99% confidence interval for population mean=3±t(0.01/2, 28)*7.32/sqrt(29)=3±2.76*7.32/sqrt(29)=3±3.75=(-0.75,6.75)

(c) here we use t-test with

null hypothesis H0:<=7.3 and

alternate hypothesis Ha:>7.3

statistic t=(-)/(sd/sqrt(n))=(3-7.3)/7.32/sqrt(29))=-0.11

since the one tailed critical at 10% level of significance is  t(0.1, 28)=1.31 is more than calculated t=-0.11, so we fail to reject H0 (or accept H0) and conclude that claim is not ture/right.

s.n.	change(x)
1	1.7
2	0.7
3	-0.1
4	-0.7
5	-3.5
6	14.9
7	3.5
8	17.1
9	-7.6
10	1.6
11	11.7
12	6.1
13	1.1
14	-4
15	20.9
16	-9.3
17	2.1
18	1.4
19	-0.3
20	-3.7
21	-1.4
22	-0.8
23	2.4
24	12.6
25	1.9
26	3.9
27	0.1
28	15.4
29	-0.7
n=	29
mean=	3.00
sd=	7.32

sum=29