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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%

Short Answer

Expert verified

The critical value \({z_{\frac{\alpha }{2}}}\)for 99% level of confidence is 2.58 (table 2.575).

Step by step solution

01

Given information

The level of significance is 99%.

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% confidence level,

\(\begin{aligned}{c}\alpha = 0.01\\1 - \frac{\alpha }{2} = 0.995\end{aligned}\)

To find the z score corresponding the area 0.9950,

In the standard normal table for positive z score, find the value closest to 0.9950, which is 0.9951, corresponding row value is 2.5, and column values is 0.08 this corresponds to the z-score of 2.58.

Therefore, the critical value for 99% level of significance is 2.58.

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Most popular questions from this chapter

In Exercises 5–8, use the given information to find the number of degrees of freedom, the critical values χL2andχR2, and the confidence interval estimate of σ. The samples are from Appendix B and it is reasonable to assume that a simple random sample has been selected from a population with a normal distribution.

Weights of Pennies 95% confidence;n= 37,s= 0.01648 g.

Sample Size. In Exercises 29–36, find the sample size required to estimate the population mean.

Mean Grade-Point Average Assume that all grade-point averages are to be standardized on a scale between 0 and 4. How many grade-point averages must be obtained so that the sample mean is within 0.01 of the population mean? Assume that a 95% confidence level is desired. If we use the range rule of thumb, we can estimate σ to be,

σ=range4=4-04=1

Does the sample size seem practical?

Replacement

Why does the bootstrap method require sampling with replacement? What would happen if we used the methods of this section but sampled without replacement?

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.

Selection Before its clinical trials were discontinued, the Genetics & IVF Institute conducted a clinical trial of the XSORT method designed to increase the probability of conceiving a girl and, among the 945 babies born to parents using the XSORT method, there were 879 girls. The YSORT method was designed to increase the probability of conceiving a boy and, among the 291 babies born to parents using the YSORT method, there were 239 boys. Construct the two 95% confidence interval estimates of the percentages of success. Compare the results. What do you conclude?

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