Question

The life in hours of a biomedical device under development in the laboratory is known to be approximately normally distributed. A random sample of 15 devices is selected and found to have an average life of 5617.8 hours and a sample standard deviation of 234.5 hours.

a) Test the hypothesis that the true mean life of a biomedical device is greater than 500 using the P - value approach.

b) Construct a 95% lower confidence bound on the mean.

c) Use the confidence bound found in part (b) to test the hypothesis.

Answer 1

a) The evidence suggest the true mean life of a biomedical device > 5500

b) (5487.94, +∞)

c) There is evidence to support that the mean is equal to 5500

Explanation:

Here we have

(a) To test the hypothesis we have the claim that the mean life of biomedical devices > 5500

Therefore, we put the null Hypothesis which is the proposition that a difference does not exist. That is

H₀: μ = 5500

Therefore, the alternative hypothesis will be

Hₐ: μ > 5500

The test statistic is then found by;

`t=\frac{\bar{x}-\mu }{(s )/(√(n))} =(5617.8-5500 )/((234.5 )/(√(15))) \approx 1.95`

The P value from the T table at df = n - 1 = 15 - 1 = 14 is

0.025 < P < 0.05

The P value is given as 0.036 from the T distribution at 14 derees of freedom df

Therefore, since P < α or 0.05, we reject the null hypothesis. That is we fail to reject Hₐ: μ > 5500. The evidence suggest the true mean life of a biomedical device > 5500

(b) The 95% confidence interval is given as

`CI=\bar{x}\pm t_\alpha (s)/(√(n))`

Which gives t
`_\alpha` =
`\pm`2.145

`CI=5617.8\pm 2.145 * (234.5)/(√(15))`

The confidence interval of the lower bound on the mean is then

(5487.94, +∞)

c) From the above result, we find that the mean of 5500 is contained in the range of the lower confidence on the mean. We can therefore, accept the null hypothesis. That is there is evidence to support that the mean is equal to 5500.