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Finding Critical Values. In Exercises 5–8, find the critical value \({{\rm{z}}_{{{\rm{\alpha }} \mathord{\left/

{\vphantom {{\rm{\alpha }} {\rm{2}}}} \right.

\kern-\nulldelimiterspace} {\rm{2}}}}}\)that corresponds to the given confidence level.

99.5%

Short Answer

Expert verified

The critical value \({z_{\frac{\alpha }{2}}}\)for 99.5% level of confidence is 2.81.

Step by step solution

01

Given information

The level of significance is 99.5%.

02

Describe the concept of critical value

A critical value is a point on the test distribution that is compared to the test statistics to determine whether to reject the null hypothesis. It is denoted by \({z_{\frac{\alpha }{2}}}\)which is equal to z score within the area of \(\frac{\alpha }{2}\)in the right tail of the standard normal distribution for\(\alpha \) level of significance.

03

Find the critical value

When finding a critical value \({z_{\frac{\alpha }{2}}}\)for a particular value of \(\alpha \), note that \(\frac{\alpha }{2}\) is the cumulative area to the right of\({z_{\frac{\alpha }{2}}}\)which implies that the cumulative area to the left of \({z_{\frac{\alpha }{2}}}\) must be\(1 - \frac{\alpha }{2}\).

Here, for 99.5% confidence level,

\(\begin{aligned}{c}\alpha = 0.005\\1 - \frac{\alpha }{2} = 0.9975\end{aligned}\)

To find the z score corresponding the area 0.9975,

In the standard normal table for positive z score, find the value 0.9975, corresponding row value is 2.8, and column values is 0.01, which corresponds to the z-score of 2.81.

Therefore, the critical value for 99.5% level of significance is 2.81.

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Most popular questions from this chapter

In Exercises 5–8, use the relatively small number of given bootstrap samples to construct the confidence interval. Freshman 15: Here is a sample of amounts of weight change (kg) of college students in their freshman year (from Data Set 6 “Freshman 15” in Appendix B): 11, 3, 0, -2, where -2 represents a loss of 2 kg and positive values represent weight gained. Here are ten bootstrap samples: {11, 11, 11, 0}, {11, -2, 0, 11}, {11, -2, 3, 0}, {3, -2, 0, 11}, {0, 0, 0, 3}, {3, -2, 3, -2}, {11, 3, -2, 0}, { -2, 3, -2, 3}, { -2, 0, -2, 3}, {3, 11, 11, 11}. a. Using only the ten given bootstrap samples, construct an 80% confidence interval estimate of the mean weight change for the population. b. Using only the ten given bootstrap samples, construct an 80% confidence interval estimate of the standard deviation of the weight changes for the population.

Formats of Confidence Intervals.

In Exercises 9–12, express the confidence interval using the indicated format. (The confidence intervals are based on the proportions of red, orange, yellow, and blue M&Ms in Data Set 27 “M&M Weights” in Appendix B.)

Red M&Ms Express 0.0434 < p < 0.217 in the form ofp+E

Finding Sample Size Instead of using Table 7-2 for determining the sample size required to estimate a population standard deviation s, the following formula can be used

n=12zα2d2

where corresponds to the confidence level and d is the decimal form of the percentage error. For example, to be 95% confident that s is within 15% of the value of s, use zα2=1.96 and d = 0.15 to get a sample size of n = 86. Find the sample size required to estimate s, assuming that we want 98% confidence that s is within 15% of σ.

Critical Thinking. In Exercises 17–28, use the data and confidence level to construct a

confidence interval estimate of p, then address the given question. Fast Food AccuracyIn a study of the accuracy of fast food drive-through orders, Burger King had 264 accurate orders and 54 that were not accurate (based on data from QSRmagazine).

a.Construct a 99% confidence interval estimate of the percentageof orders that are not accurate.

b.Compare the result from part (a) to this 99% confidence interval for the percentage of orders that are not accurate at Wendy’s: 6.2%<p< 15.9%. What do you conclude?

Genes Samples of DNA are collected, and the four DNA bases of A, G, C, and T are codedas 1, 2, 3, and 4, respectively. The results are listed below. Construct a 95% confidence intervalestimate of the mean. What is the practical use of the confidence interval?

2 2 1 4 3 3 3 3 4 1

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