Answered

Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Discover precise answers to your questions from a wide range of experts on our user-friendly Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.

A standard weight known to weigh 10 grams. Some suspect bias in weights due to manufacturing process. To assess the accuracy of a laboratory scale,that weight is weighed repeatedly. The scale readings are normally distributed with unknown mean. The population standard deviation of the scale readings is known to be 0.0003 gram.
(a) The weight is weighed FIVE times. The mean result is 10.0044 grams. Give a 98% confidence interval for the mean weight. Please keep 5 decimals in your answer.
(
,
)

(b) How many measurements must be averaged to get a margin of error of +/- 0.0001 with 98% confidence? Your answer must be a whole number.

Sagot :

Answer:

a) The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams

b) 49 measurements are needed.

Step-by-step explanation:

Question a:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1 - 0.98}{2} = 0.01[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1 - \alpha[/tex].

That is z with a pvalue of [tex]1 - 0.01 = 0.99[/tex], so Z = 2.327.

Now, find the margin of error M as such

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

[tex]M = 2.327\frac{0.0003}{\sqrt{5}} = 0.00031[/tex]

The lower end of the interval is the sample mean subtracted by M. So it is 10.0044 - 0.00031 = 10.00409 grams

The upper end of the interval is the sample mean added to M. So it is 10 + 0.00031 = 10.00471 grams

The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams.

(b) How many measurements must be averaged to get a margin of error of +/- 0.0001 with 98% confidence?

We have to find n for which M = 0.0001. So

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

[tex]0.0001 = 2.327\frac{0.0003}{\sqrt{n}}[/tex]

[tex]0.0001\sqrt{n} = 2.327*0.0003[/tex]

[tex]\sqrt{n} = \frac{2.327*0.0003}{0.0001}[/tex]

[tex](\sqrt{n})^2 = (\frac{2.327*0.0003}{0.0001})^2[/tex]

[tex]n = 48.73[/tex]

Rounding up

49 measurements are needed.

We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.