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8090 [49]
3 years ago
15

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.
Mathematics
1 answer:
Anton [14]3 years ago
6 0

Answer:

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

Step-by-step 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 }{\frac{s }{\sqrt{n}}} =\frac{5617.8-5500 }{\frac{234.5 }{\sqrt{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 \frac{s}{\sqrt{n}}

Which gives    t_\alpha =  \pm2.145

CI=5617.8\pm 2.145 \times \frac{234.5}{\sqrt{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.

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Answer:

The required recursive formula is

f(n)= 500\times(1.4)^{n-1}\\

Step-by-step explanation:

Mohamed decided to track the number of leaves on the tree in his backyard each year.

The first year there were 500 leaves

Year \: 1 = 500

Each year thereafter the number of leaves was 40% more than the year before so that means

Year \: 2 = 500(1+0.40) = 500\times 1.4\\

For the third year the number of leaves increase 40% than the year before so that means

Year \: 3 = 500\times 1.4(1+0.40) = 500 \times 1.4^{2}\\

Similarly for fourth year,

Year \: 4 = 500\times 1.4^{2}(1+0.40) = 500\times 1.4^{3}\\

So we can clearly see the pattern here

Let f(n) be the number of leaves on the tree in Mohameds back yard in the nth year since he started tracking it then general recursive formula is

f(n)= 500\times(1.4)^{n-1}\\

This is the required recursive formula to find the number of leaves for the nth year.

Bonus:

Lets find out the number of leaves in the 10th year,

f(10)= 500\times(1.4)^{10-1}\\\\f(10)= 500\times(1.4)^{9}\\\\f(10)= 500\times20.66\\\\f(10)= 10330

So there will be 10330 leaves in the 10th year.

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