A sample of 24 paired observations generates the following data: 1formula155.mml

Question

A sample of 24 paired observations generates the following data: 1formula155.mml = 0.7 and 1formula156.mml = 3.2. Assume a normal distribution. Use Table 2. a. Construct the 95% confidence interval for the mean difference D. (Round all intermediate calculations to at least 4 decimal places and final answers to 2 decimal places.) Confidence interval is to . b. Using the confidence interval, test whether the mean difference differs from zero. There is no evidence that the mean difference differs from zero. There is evidence that the mean difference differs from zero.

Explanation / Answer

Solution:

a) It is given that, xd = 0.7, Sd = 3.2 and n = 24
The degrees of freedom is,
Df = n-1 = 24-1 = 23
The critical value using excel function as shown below:
t/2, n-1 = t0.05/2, 24-1
= T.INV.2T(0.05,23) = ±2.0687
The 95% confidence interval for d is

xd ± t/2,n-1(Sd/n) = 0.7 ± (2.0687)(3.2/24)
= 0.7 ± (2.0687) (0.6532)
= (-0.65, 2.05)
The 95% confidence intervals id -0.65 < d < 2.05

b) The null and alternative hypothesis can be defined as follows:
H0: d = 0
H1: d 0
From part (a), the 95% confidence interval is -0.65 < d < 2.05
Here, the confifence intervl not include zero, the null hypothesis should be rejected.
Therfore, it can be concluded that there is sufficient evidence to support that the mean difference differes from xero

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