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Chapter 7: Problem 6

On the basis of extensive tests, the yield point of a particular type of mild steel-reinforcing bar is known to be normally distributed with \(\sigma=100\). The composition of the bar has been slightly modified, but the modification is not believed to have affected either the normality or the value of \(\sigma\). a. Assuming this to be the case, if a sample of 25 modified bars resulted in a sample average yield point of \(8439 \mathrm{lb}\), compute a \(90 \%\) CI for the true average yield point of the modified bar. b. How would you modify the interval in part (a) to obtain a confidence level of \(92 \%\) ?

Short Answer

Expert verified
a. 90% CI is (8406.1, 8471.9); b. 92% CI is (8404, 8474).

Step by step solution

01

Identify Given Values

We are given that the population standard deviation \(\sigma\) is 100. The sample mean \(\bar{x}\) is 8439, and the sample size \(n\) is 25. In part (a), we are tasked with calculating a 90% confidence interval for the true average yield point.
02

Find the Z-value for 90% Confidence Level

For a 90% confidence interval, we need the critical Z-value. Since the confidence level is two-tailed, we divide the significance level, \(\alpha = 1 - 0.9 = 0.1\), by 2, giving \(\alpha/2 = 0.05\). The corresponding Z-value that leaves 0.05 in each tail is approximately 1.645.
03

Calculate the Standard Error (SE)

The standard error (SE) is calculated using the formula: \(SE = \frac{\sigma}{\sqrt{n}}\). Substituting the given values, \(SE = \frac{100}{\sqrt{25}} = \frac{100}{5} = 20\).
04

Compute the Confidence Interval (CI) for 90%

The confidence interval is calculated using the formula: \( \bar{x} \pm Z \times SE \). Substituting the values, we get: \( 8439 \pm 1.645 \times 20 \). Calculate the margin of error, which is \(1.645 \times 20 = 32.9\). Therefore, the 90% confidence interval is \(8439 \pm 32.9\), which is \((8406.1, 8471.9)\).
05

Find the Z-value for 92% Confidence Level

For a 92% confidence interval, calculate the corresponding Z-value. The significance level is \(\alpha = 1 - 0.92 = 0.08\), and \(\alpha/2 = 0.04\). Consult a Z-table or use a calculator to find that the Z-value is approximately 1.75.
06

Adjust the Confidence Interval to 92%

Using the new Z-value of 1.75, recalculate the confidence interval: \( \bar{x} \pm Z \times SE = 8439 \pm 1.75 \times 20 \). The margin of error is now \(1.75 \times 20 = 35\). Thus, the 92% confidence interval is \((8404, 8474)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Distribution
In statistics, a normal distribution is a common occurrence. Most data points in a natural setting fall into this pattern, resembling a symmetrical bell curve. The highest point on this curve represents the mean, mode, and median, all located at the center.
  • The spread of the data is controlled by two aspects: the mean and standard deviation.
  • A normal distribution is denoted by the symbol "N" followed by its mean and standard deviation in parentheses, like this: \( N(\mu, \sigma) \).
Normal distributions are important because many tests and confidence intervals assume that the data follows a normal distribution. For this problem, it's crucial to know that the yield point data of steel-reinforcing bars are assumed to be normally distributed.
Population Standard Deviation
The population standard deviation, represented with the Greek letter \( \sigma \), measures the amount of variation or dispersion in a population set. It helps in understanding how spread out the population data is.
  • In this example, we have a known standard deviation of 100 for the yield point of the bars.
  • A smaller \( \sigma \) means data points are closer to the mean, whereas a larger \( \sigma \) indicates more spread.
Knowing \( \sigma \) allows us to use the normal distribution and compute the standard error when creating confidence intervals, which are fundamental in estimating unknown population parameters with a certain level of confidence.
Sample Size
Sample size, symbolized by \( n \), is critical in statistical analysis. It represents the number of observations in a sample and has a significant impact on the reliability of statistical results.
  • In our problem, \( n = 25 \) indicates that 25 modified steel bars were selected to calculate the average yield point.
  • Larger sample sizes tend to provide more accurate estimates of the population parameter, reducing the standard error and thus narrowing the confidence interval.
Choosing the right sample size is vital as it affects the study's conclusions and the confidence interval's precision. A balance is needed because larger samples require more resources, but provide better accuracy.
Z-value
The Z-value, also known as the standard score, shows how many standard deviations a data point is from the mean. It's crucial for calculating confidence intervals, as it helps determine the cutoff points for a given level of confidence.
  • In our calculation, different Z-values are used for 90% and 92% confidence intervals, specifically 1.645 and 1.75, respectively.
  • Z-values are found using standard normal distribution tables or calculators. The two-tailed approach divides the total significance level (alpha) by 2 to find these values.
Knowing the Z-value enables us to compute the margin of error. This is added to and subtracted from the sample mean to establish the upper and lower limits of the confidence interval.

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Most popular questions from this chapter

A random sample of 539 households from a certain midwestern city was selected, and it was determined that 133 of these households owned at least one firearm ("The Social Determinants of Gun Ownership: Self-Protection in an Urban Environment," Criminology, 1997: 629-640). Using a 95\% confidence level, calculate a lower confidence bound for the proportion of all households in this city that own at least one firearm.

Determine the following: a. The 95 th percentile of the chi-squared distribution with \(v=10\) b. The 5 th percentile of the chi-squared distribution with \(v=10\) c. \(P\left(10.98 \leq \chi^{2} \leq 36.78\right)\), where \(\chi^{2}\) is a chi- squared rv with \(v=22\) d. \(P\left(\chi^{2}<14.611\right.\) or \(\left.\chi^{2}>37.652\right)\), where \(\chi^{2}\) is a chisquared rv with \(v=25\)

A CI is desired for the true average stray-load loss \(\mu\) (watts) for a certain type of induction motor when the line current is held at 10 amps for a speed of \(1500 \mathrm{rpm}\). Assume that strayload loss is normally distributed with \(\sigma=3.0\). a. Compute a \(95 \%\) CI for \(\mu\) when \(n=25\) and \(\bar{x}=58.3\). b. Compute a \(95 \%\) CI for \(\mu\) when \(n=100\) and \(\bar{x}=58.3\). c. Compute a \(99 \%\) CI for \(\mu\) when \(n=100\) and \(\bar{x}=58.3\). d. Compute an \(82 \%\) CI for \(\mu\) when \(n=100\) and \(\bar{x}=58.3\). e. How large must \(n\) be if the width of the \(99 \%\) interval for \(\mu\) is to be \(1.0\) ?

Determine the \(t\) critical value that will capture the desired \(t\) curve area in each of the following cases: a. Central area \(=.95\), df \(=10\) b. Central area \(=.95\), df \(=20\) c. Central area \(=.99\), df \(=20\) d. Central area \(=.99\), df \(=50\) e. Upper-tail area \(=.01\), df \(=25\) f. Lower-tail area \(=.025\), df \(=5\)

Suppose that a random sample of 50 bottles of a particular brand of cough syrup is selected and the alcohol content of each bottle is determined. Let \(\mu\) denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the resulting \(95 \%\) confidence interval is \((7.8,9.4)\). a. Would a \(90 \%\) confidence interval calculated from this same sample have been narrower or wider than the given interval? Explain your reasoning. b. Consider the following statement: There is a \(95 \%\) chance that \(\mu\) is between \(7.8\) and \(9.4\). Is this statement correct? Why or why not? c. Consider the following statement: We can be highly confident that \(95 \%\) of all bottles of this type of cough syrup have an alcohol content that is between \(7.8\) and \(9.4\). Is this statement correct? Why or why not? d. Consider the following statement: If the process of selecting a sample of size 50 and then computing the corresponding \(95 \%\) interval is repeated 100 times, 95 of the resulting intervals will include \(\mu\). Is this statement correct? Why or why not?
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