Download Explain why it would not be reasonable to use estimation after a

Explain why it would not be reasonable to use estimation after a hypothesis test for which the decision was fail to reject
Ho.
The Ho could not be rejected implies that the difference between the sample statistic and the population parameter was
not significant. This means the treatment had no effect. Therefore, estimation can’t be used.
A researcher obtains a sample from an unknown population and computes a sample mean of M =43 with a standard
deviation of s = 6. a. If the sample has n= 16 scores, compute an 80% confidence interval to estimate the unknown
population mean. b. If the sample has n =36 scores, computer an 80% confidence interval to estimate the unknown
population mean. c. Comparing your answers for a and b, describe how sample size influences the width of a confidence
interval.
(a)
Data:
n
x-bar
σ or s
%
16
43
6
80
Standard Error, SE = s/√n =
1.5000
Degrees of freedom = 15
t- score =
1.3406
Width of the confidence interval = t * SE =
(b)
2.0109
Lower Limit of the confidence interval = x-bar - width =
40.9891
Upper Limit of the confidence interval = x-bar + width =
45.0109
The confidence interval is
]
Data:
[ 40.9891
45.0109
n
x-bar
σ or s
%
36
43
6
80
Standard Error, SE = s/√n =
1.0000
Degrees of freedom = 35
t- score =
1.3062
Width of the confidence interval = t * SE =
(c)
1.3062
Lower Limit of the confidence interval = x-bar - width =
41.6938
Upper Limit of the confidence interval = x-bar + width =
44.3062
The confidence interval is
]
[ 41.6938
44.3062
From (a) and (b) we can infer that as the sample size increases, the width of the confidence interval
decreases.
Standardized measures seem to indicate that the average level of anxiety has increased gradually over the past 50 years
(Twenge, 2000). In the 1950s, the average score on the Child Manifest Anxiety Scale was u = 15.1. A sample of n = 16 of
today’s children produces a mean score of M = 23.3 with SS = 240. a. Based on the sample, make a point estimate of the
population mean anxiety score for today’s children. b. Make a 90% confidence interval estimate of today’s population
mean.
(a)
The point estimate is 23.3
(b)
s^2 = SS/(n - 1) = 240/(16 - 1) = 16, and so s = 4
Data:
n
x-bar
σ or s
%
16
23.3
4
90
Standard Error, SE = s/√n =
1.0000
Degrees of freedom =
15
t- score =
1.7531
Width of the confidence interval = t * SE =
1.7531
Lower Limit of the confidence interval = x-bar - width =
21.5469
Upper Limit of the confidence interval = x-bar + width =
25.0531
The confidence interval is
]
[
21.5469
25.0531