/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 111 If 36 measurements of the specif... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 111

If 36 measurements of the specific gravity of aluminum had a mean of 2.705 and a standard deviation of .028 , construct a \(98 \%\) confidence interval for the actual specific gravity of aluminum.

Short Answer

Expert verified
Answer: The 98% confidence interval for the true specific gravity of aluminum based on the given sample is (2.69298, 2.71702).

Step by step solution

01

Identify the given information

Sample mean (x虆) = 2.705, Standard deviation (蟽) = 0.028, Sample size (n) = 36, and Confidence level = 98%.
02

Find the critical value for 98% confidence interval

First, we need to find the critical value for the given confidence level. For a 98% confidence interval, there is a 2% chance that the true mean is not within the interval. Since we are using a two-tailed distribution (as we are estimating an upper and lower bound for the confidence interval), we need to find the critical value for the 99% confidence level (1 - (2% / 2)). Using a Z-table or calculator, we find that the critical value (Z) for a 99% confidence level is approximately 2.576.
03

Compute the standard error of the mean

The standard error of the mean is calculated using the formula: Standard Error (SE) = \(\frac{蟽}{\sqrt{n}}\) Plug in the given values: SE = \(\frac{0.028}{\sqrt{36}}\) = \(\frac{0.028}{6}\) = 0.00466667
04

Calculate the margin of error

Margin of error (ME) is calculated using the formula: ME = Z * SE Plug in the values: ME = 2.576 * 0.00466667 鈮� 0.01202
05

Construct the 98% confidence interval

The confidence interval is calculated using the formula: Confidence Interval (CI) = (x虆 - ME, x虆 + ME) Plug in the values: CI = (2.705 - 0.01202, 2.705 + 0.01202) = (2.69298, 2.71702) Therefore, we can say with 98% confidence that the true specific gravity of aluminum lies within the interval (2.69298, 2.71702).

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Most popular questions from this chapter

In a study of the relationship between birth order and college success, an investigator found that 126 in a sample of 180 college graduates were firstborn or only children. In a sample of 100 nongraduates of comparable age and socioeconomic background, the number of firstborn or only children was \(54 .\) Estimate the difference between the proportions of firstborn or only children in the two populations from which these samples were drawn. Use a \(90 \%\) confidence interval and interpret your results.

Americans are becoming more conscious of the importance of good nutrition, and some researchers believe that we may be altering our diets to include less red meat and more fruits and vegetables. To test this theory, a researcher decides to select hospital nutritional records for subjects surveyed 10 years ago and to compare the average amount of beef consumed per year to the amounts consumed by an equal number of subjects she will interview this year. She knows that the amount of beef consumed annually by Americans ranges from 0 to approximately 104 pounds. How many subjects should the researcher select for each group if she wishes to estimate the difference in the average annual per- capita beef consumption correct to within 5 pounds with \(99 \%\) confidence?

Independent random samples of \(n_{1}=800\) and \(n_{2}=640\) observations were selected from binomial populations 1 and \(2,\) and \(x_{1}=337\) and \(x_{2}=374\) successes were observed. a. Find a \(90 \%\) confidence interval for the difference \(\left(p_{1}-p_{2}\right)\) in the two population proportions. Interpret the interval. b. What assumptions must you make for the confidence interval to be valid? Are these assumptions met?

Suppose the number of successes observed in \(n=500\) trials of a binomial experiment is \(27 .\) Find a \(95 \%\) confidence interval for \(p\). Why is the confidence interval narrower than the confidence interval in Exercise \(8.25 ?\)

Don't Americans know that eating pizza and french fries leads to being overweight? In the same American Demographics article referenced in Exercise \(8.98,\) a survey of women who are the main meal preparers in their households reported these results: \(\cdot$$90 \%\) know that obesity causes health problems. \(\cdot$$80 \%\) know that high fat intake may lead to health problems. \(\cdot$$86 \%\) know that cholesterol is a health problem. \(\cdot$$88 \%\) know that sodium may have negative effects on health. a. Suppose that this survey was based on a random sample of 750 women. How accurate do you expect the percentages given above to be in estimating the actual population percentages? (HINT: If these are the only four percentages for which you need a margin of error, a conservative estimate for \(p\) is \(p \approx .80 .)\) b. If you want to decrease your sampling error to \(\pm 1 \%,\) how large a sample should you take?
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