*Please show work For exercise 12.2- Assume that the paired sample data are simp

Question

*Please show work

For exercise 12.2- Assume that the paired sample data are simple random samples and that the differences have a distribution that is approximately normal.

12.2 – Body Temperatures. Use the sample data from exercise 12.1 to test the claim that there is no difference between body temperatures measured at 8 A.M. and at 12 A.M. use a 0.05 significance level. Follow the steps below:

a) Identify the null & alternative hypothesis.

b) Find the P-Value. Is the P-value greater or less than the significance level of ? = 0.05? Do we reject or fail to reject the null hypothesis?

Sample Data:

Explanation / Answer

The data given

X( temp at 8AM): 98 97.0 98.6 97.4

Y (temp at 12 AM): 98 97.6 98.8 98.0

Let the null hypothesis be H0: µx = µy i.e. there is no significant difference between the mean body temperature measured at 8AM and 12 AM against the alternative hypothesis H1: µx    µy

At 8 AM the mean X = 97.75 At 12 AM the mean Y= 98.1

S2 = 1/(n1+n2-2)[ (x - X)2 + ( y - Y)2 ]

= 1/(4+4-2) [1.47 + 0.76]

= 0.371667

The test Statistic t under the nul hypothesis is = (X - Y) / sqrt ( S2(1/n1 +1/n2)

   = (97.75 - 98.1) / 0.371667 = 0.941

The tabulated value of t for (4+4-2) df = 6 df is 2.45 for 0.05 level of significance

So the hypothesis of equality is accepted.

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