Question

Some types of nerve cells have the ability to regenerate a part of the cell that has been amputated. In a study of this process in lemurs, researchers cut nerves emanating from the spinal cord of a sample of 31animals and measured the content of creatine phosphate (CP). They found a sample mean of 0.213 mg/g, with a standard deviation of 0.094 mg/g.In healthy cells, the CP content is 0.15 mg/g. Assuming all relevant assumptions hold,conduct the appropriate test to determine if CP content in regenerating nerve cells differs from this amount in this population of lemurs.Given: t∗df=2.042

Answer 1

The provided sample mean is and the sample standard deviation is , and the sample size is n=31.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:

Ha:

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a two-tailed test is tc​=2.042.

The rejection region for this two-tailed test is R={t:∣t∣>2.042}

(3) Test Statistics

The t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that ∣t∣=3.732>tc​=2.042, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0008, and since p=0.0008<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is different than 0.15, at the 0.05 significance level.

Graphically

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