In: Statistics and Probability
A scientist claims that pneumonia is associated with weight loss in mice. The table shows the weights (in grams) of six mice before infection and two days after the infection. At α=0.01 is there enough evidence to support the scientist’s claim? (Assume the samples are random and dependent, and the populations are normally distributed). Hint: We are testing the claim that the weight before pneumonia will be greater than the weight after.
Weight (before) 19.8 20.6 20.3 22.1 23.4 23.6
Weight (after) 18.4 19.6 19.6 20.7 22.2 23.0
a) State the null and alternative hypothesis
b) List the following to two decimal places a. ?̅ b. ??
c) Calculate the test statistic (two decimal places)
d) Calculate the p-value (three decimal places)
e) Reject or fail to reject the null hypothesis? Why?
f) State your conclusion in context of the problem
The data is:
Pair | Before | After | Difference |
1 | 19.8 | 18.4 | 1.4 |
2 | 20.6 | 19.6 | 1 |
3 | 20.3 | 19.6 | 0.7 |
4 | 22.1 | 20.7 | 1.4 |
5 | 23.4 | 22.2 | 1.2 |
6 | 23.6 | 23 | 0.6 |
Paired Sample t test |
For the score differences we have, mean is Dˉ=1.05, the sample
standard deviation is sD=0.345, and the sample size is n=6. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: μD =0 Ha: μD >0 This corresponds to a Right-tailed test, for which a t-test for two paired samples be used. (2a) Critical Value Based on the information provided, the significance level is α=0.01, and the degree of freedom is n-1=6-1=5. Therefore the critical value for this Right-tailed test is tc=3.3649. This can be found by either using excel or the t distribution table. (2b) Rejection Region The rejection region for this Right-tailed test is t>3.3649 (3)Test Statistics The t-statistic is computed as follows: (4) The p-value The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case, the p-value is 0.0003 (5) The Decision about the null hypothesis (a) Using traditional method Since it is observed that t=7.4558 > tc=3.3649, it is then concluded that the null hypothesis is rejected. (b) Using p-value method Using the P-value approach: The p-value is p=0.0003, and since p=0.0003≤0.01, it is concluded that the null hypothesis is rejected. (6) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ1 is greater than μ2, at the 0.01 significance level. Hence there is sufficient evidence to support the claim that the weight before pneumonia will be greater than the weight after.. |
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