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.

98%

Short Answer

Expert verified

The critical value \({z_{\frac{\alpha }{2}}}\)for 98% level of confidence is 2.33.

Step by step solution

01

Given information

The level of significance is 98%.

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 98% confidence level,

\(\begin{array}{c}\alpha = 0.02\\1 - \frac{\alpha }{2} = 0.99\end{array}\)

To find the z score corresponding to the area 0.9900,

In the standard normal table for positive z score, find the value closest to 0.9900, which is 0.9901, corresponding row value is 2.3 and column values is 0.03which corresponds to the z-score of 2.33.

Therefore, the critical value for 98% level of significance is 2.33.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Exercises 9–16, assume that each sample is a simple

random sample obtained from a population with a normal distribution.

Body Temperature Data Set 3 “Body Temperatures” in Appendix B includes a sample of106 body temperatures having a mean of 98.20°F and a standard deviation of 0.62°F (for day 2at 12 AM). Construct a 95%confidence interval estimate of the standard deviation of the bodytemperatures for the entire 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.)

Orange M&Ms Express 0.179 < p < 0.321 in the form of p^±E.

Use the given data to find the minimum sample size required to estimate a population proportion or percentage. Lefties: Find the sample size needed to estimate the percentage of California residents who are left-handed. Use a margin of error of three percentage points, and use a confidence level of 99%.

a. Assume that p^andq^are unknown.

b. Assume that based on prior studies, about 10% of Californians are left-handed.

c. How do the results from parts (a) and (b) change if the entire United States is used instead of California?

In Exercises 5–8, use the relatively small number of given bootstrap samples to construct the confidence interval.Seating Choice In a 3M Privacy Filters poll, respondents were asked to identify their favourite seat when they fly, and the results include these responses: window, window, other, other. Letting “window” = 1 and letting “other” = 0, here are ten bootstrap samples for those responses: {0, 0, 0, 0}, {0, 1, 0, 0}, {0, 1, 0, 1}, {0, 0, 1, 0}, {1, 1, 1, 0}, {0, 1, 1, 0}, {1, 0, 0, 1}, {0, 1, 1, 1}, {1, 0, 1, 0}, {1, 0, 0, 1}. Using only the ten given bootstrap samples, construct an 80% confidence interval estimate of the proportion of respondents who indicated their favourite seat is “window.”

Confidence Intervals. In Exercises 9–24, construct the confidence interval estimate of the mean.

Flight Arrivals Listed below are arrival delays (minutes) of randomly selected American Airlines flights from New York (JFK) to Los Angeles (LAX). Negative numbers correspond to flights that arrived before the scheduled arrival time. Use a 95% confidence interval. How good is the on-time performance?

-5 -32 -13 -9 -19 49 -30 -23 14 -21 -32 11

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free