Part A) The one-sample t statistic for testing H 0 : = 40 H a : 40 from a sample

Question

Part A)

The one-sample t statistic for testing

H0: = 40

Ha: 40

from a sample of n = 15 observations has the value t = 2.72.

(a)

What are the degrees of freedom for t?

(b)

Locate the two critical values t* from the Table D that bracket t.

< t <

(c)

Between what two values does the P-value of the test fall?

0.005 < P < 0.010.01 < P < 0.02    0.02 < P < 0.040.04 < P < 0.050.05 < P < 0.1

(d)

Is the value t = 2.72 statistically significant at the 5% level?

YesNo    

Is it significant at the 1% level?

YesNo    

(e)

If you have software available, find the exact P-value. (Round your answer to four decimal places.)

Part B)

Self-efficacy is a general concept that measures how well we think we can control different situations. A multimedia program designed to improve dietary behavior among low-income women was evaluated by comparing women who were randomly assigned to intervention and control groups. Participants were asked, "How sure are you that you can eat foods low in fat over the next month?" The response was measured on a five-point scale with 1 corresponding to "not sure at all" and 5 corresponding to "very sure." Here is a summary of the self-efficacy scores obtained about 2 months after the intervention:

x

1.12

What is the df=

x

1.12

Explanation / Answer

Part a)

a) The degrees of freedom for the test would be computed as:

DF = n- 1 = 15 -1 = 14

Therefore there are 14 degrees of freedom for the test here.

b) Here we are not given the level of significance for the test, assuming it to be 5%, we get from the t-distribution tables that:

P( - 2.145 < t14 < 2.145 ) = 0.95

Therefore - 2.145 , 2.145 are the critical values here.

c) Now here we are given the test statistic value as 2.72, therefore for this two tailed test, the p-value here is computed from the t-distribution tables as:

p = 2P(t14 > 2.72 ) = 2*0.0083 = 0.0166

Therefore the correct interval for the p-value here is :

0.01 < P < 0.02

d) Now as the p-value here is 0.0166 < 0.05 therefore the test is significant at 5% level of significance.

But as the p-value is 0.0166 > 0.01, therefore the test is not significant at 1% level of significance.

e) The exact p-value is already computed above equal to 0.0166

Part b)

The degree of freedom for a two sample t test is computed as:

DF = n1 + n2 - 2 = 163 + 217 - 2 = 378

This means 378 degrees of freedom.

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