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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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If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t
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1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

The number 1052, written as a base 6 number is 4512

Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, 6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296

This implies that 6^{4} exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, 6^{3}=216 and 216\times 5=1080, 216\times 4=864. This implies that 216\times 5 exceeds 1052 and thus there can be at most 4 groups of 6^{3}.

Then,

1052-4\times6^{3}

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Now, 6^{2}=36 and 36\times 5=180, 36\times 6=216. This implies that 36\times 6 exceeds 188 and thus there can be at most 5 groups of 6^{2}.

Then,

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So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

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Then,

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So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, 6^{0}=1 and 1\times 2=2. This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

This concludes the grouping.

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So, the number 1052, written as a base 6 number is 4512.

Learn more about change of base of numbers here:

brainly.com/question/14291917

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