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Chapter 9: Problem 14

What is the difference between the critical value of \(z\) and the observed value of \(z\) ?

Short Answer

Expert verified
The difference between the critical value and the observed value of z can be calculated by subtraction (Observed value - Critical value). However, this difference doesn't always have a meaningful interpretation, as they represent different concepts in statistics 鈥� the critical value determines what is 'significantly' far from the mean, while the observed value shows how far your data's mean is from the population mean in terms of standard deviations.

Step by step solution

01

Understand the Critical Value

The critical value in a z-distribution is the z-score which segregates the area of the z-distribution such that the area of one side quantifies the alpha level (伪), which is the probability of rejecting the null hypothesis when the null hypothesis is true. The typical values of code alpha are 0.05 or 0.01; hence, the critical z-scores for these alpha levels are approximately 卤1.96 and 卤2.58 (for a two-tailed test).
02

Understand the Observed Value

The observed value of z (also known as the test statistic) is calculated by the formula: \(z = (X虅 - 渭鈧�)/(蟽/鈭歯)\), where X虅 is the sample mean, 渭鈧� is the population mean, 蟽 is the population standard deviation, and n is the sample size. This z-score tells us how many standard deviations our sample mean is away from the population mean.
03

Computing the Difference

Once the critical and observed z-scores have been established, the difference between the two can be calculated using simple subtraction: Difference = Observed value - Critical value. It is important to note that calculation of difference might not always result in a meaningful value, as these two values serve different purposes in hypothesis testing 鈥� indicating the significance of results and showing how 'unusual' your data is, respectively.

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Most popular questions from this chapter

A past study claimed that adults in America spent an average of 18 hours a week on leisure activities. A researcher wanted to test this claim. She took a sample of 12 adults and asked them about the time they spend per week on leisure activities. Their responses (in hours) are as follows. \(\begin{array}{lllllllllll}13.6 & 14.0 & 24.5 & 24.6 & 22.9 & 37.7 & 14.6 & 14.5 & 21.5 & 21.0 & 17.8 & 21.4\end{array}\) Assume that the times spent on leisure activities by all adults are normally distributed. Using the \(10 \%\) significance level, can you conclude that the average amount of time spent on leisure activities has changed?

Shulman Steel Corporation makes bearings that are supplied to other companies. One of the machines makes bearings that are supposed to have a diamcter of 4 inches. The bearings that have a diameter of either more or less than 4 inches are considered defective and are discarded. When working properly, the machine does not produce more than \(7 \%\) of bearings that are defective. The quality control inspector selects a sample of 200 bearings each week and inspects them for the size of their diameters. Using the sample proportion, the quality control inspector tests the null hypothesis \(p \leq .07\) against the alternative hypothesis \(p \geq\) 07, where \(p\) is the proportion of bearings that are defective. He always uses a \(2 \%\) significance level. If the null hypothesis is rejected, the machine is stopped to make any necessary adjustments. One sample of 200 bearings taken recently contained 22 defective bearings. a. Using the \(2 \%\) significance level, will you conclude that the machine should be stopped to make necessary adjustments? b. Perform the test of part a using a \(1 \%\) significance level. Is your decision different from the one in part a?

A real estate agent claims that the mean living area of all single-family homes in his county is at most 2400 square feet. A random sample of 50 such homes selected from this county produced the mean living area of 2540 square feet and a standard deviation of 472 square feet. a. Using \(\alpha=.05\), can you conclude that the real estate agent's claim is true? What will your conclusion be if \(\alpha=.01 ?\)

According to the American Diabetes Association (www.diabetes.org), \(23.1 \%\) of Americans aged 60 years or older had diabetes in 2007. A recent random sample of 200 Americans aged 60 years or older showed that 52 of them have diabetes. Using a \(5 \%\) significance level, perform a test of hypothesis to determine if the current percentage of Americans aged 60 years or older who have diabetes is higher than that in 2007 . Use both the \(p\) -value and the critical-value approaches.

An carlier study claimed that U.S. adults spent an average of 114 minutes with their families per day. A recently taken sample of 25 adults from a city showed that they spend an average of 109 minutes per day with their families. The sample standard deviation is 11 minutes. Assume that the times spent by adults with their families have an approximately normal distribution. a. Using the \(1 \%\) significance level, test whether the mean time spent currently by all adults with their families in this city is different from 114 minutes a day. b. Suppose the probability of making a Type I error is zero. Can you make a decision for the test of part a without going through the five steps of hypothesis testing? If yes, what is your decision? Explain.
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