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Consider the following null and alternative hypotheses: $$ H_{0}: \mu=40 \quad \text { versus } \quad H_{1}: \mu \neq 40 $$ A random sample of 64 observations taken from this population produced a sample mean of \(38.4\). The population standard deviation is known to be \(6 .\) a. If this test is made at the \(2 \%\) significance level, would you reject the null hypothesis? Use the critical-value approach. b. What is the probability of making a Type I error in part a? c. Calculate the \(p\) -value for the test. Based on this \(p\) -value, would you reject the null hypothesis if \(\alpha=01\) ? What if \(\alpha=05\) ?

Short Answer

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a. We fail to reject the null hypothesis. b. The probability of making a Type I error is 2%. c. Given the calculated P-value of 0.1096, we fail to reject the null hypothesis regardless of whether our significance level is 0.01 or 0.05.

Step by step solution

01

Computing the Z-score

The first step in hypothesis testing is computing the Z-score. The Z-score can be obtained by using the formula: \( Z = \frac{{\bar{x}-\mu}}{{\sigma/\sqrt{n}}} \) where, \(\bar{x}\) = sample mean = 38.4, \(\mu\) = population mean (under null hypothesis) = 40, \(\sigma\) = population standard deviation = 6, \(n\) = number of observations = 64. Plugging these values into the formula we get \( Z = \frac{{38.4 - 40}}{{6/\sqrt{64}}} = -1.6 \).
02

Identifying Critical Z-value

This is a two-tailed test because the alternative hypothesis is \( \mu \neq 40 \). The significance level is 2%, which is split between the two tails, making 1% in each tail. Looking up the z-table for an area of 1% in the tail, we find that the critical z-value is ±2.33.
03

Comparing Z-score with Critical Z-value

Comparing our calculated Z-score with the critical Z-value, -1.6 does not fall into the critical region of ±2.33. Therefore, we fail to reject the null hypothesis.
04

Calculating probability of Type I error

A Type I error occurs when we incorrectly reject the null hypothesis when it is true. The probability of committing a Type I error is equal to the given significance level, which in this case is 0.02 or 2%.
05

Calculating the P-value

The P-value is the minimum significance level at which we would reject the null hypothesis given the observed sample data. It corresponds to the observed Z-score. For a two-tailed test, the P-value is the total probability in both tails, given the Z-score. Looking up -1.6 on the Z-table, we find the area in the left tail to be 0.0548. Since it's a two-tailed test, our P-value is \(2*0.0548 = 0.1096\).
06

Comparing P-value with Significance Level

Comparing the calculated P-value with our level of significance, 0.1096 is greater than 0.01 and 0.05. Regardless of whether our significance level is 0.01 or 0.05, we fail to reject the null hypothesis.

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Key Concepts

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

Understanding the Null Hypothesis
In statistical analysis, the null hypothesis serves as a baseline statement. It proposes that there is no effect or no difference in the context of the study. For our exercise, the null hypothesis is that the population mean (\( \mu \)) equals 40, denoted as \( H_0: \mu = 40 \). It acts as the default assumption we test against.

Hypothesis testing aims to determine whether there is enough evidence to reject this null hypothesis in favor of an alternative hypothesis, in this case, \( H_1: \mu eq 40 \). Rejecting or failing to reject the null depends on the data analysis and chosen significance level. It's important to understand that failing to reject the null doesn't prove it's true; it simply implies there's insufficient evidence against it.
Deciphering the Z-score
The Z-score is a statistical metric that tells us how many standard deviations an observation is from the mean. It acts as a tool for comparing the sampled data to an assumed population mean, allowing us to make decisions about the null hypothesis.

In our scenario, we compute the Z-score using the formula: \[ Z = \frac{{\bar{x} - \mu}}{{\sigma/\sqrt{n}}} \] where the sample mean \( \bar{x} = 38.4 \), the population mean \( \mu = 40 \), the standard deviation \( \sigma = 6 \), and our number of observations \( n = 64 \).
Plugging these numbers in, we find \( Z = -1.6 \). This tells us that the sample mean is 1.6 standard deviations less than the hypothesized population mean. Understanding the Z-score helps determine where the mean falls in relation to the null hypothesis.
Recognizing Type I Error
When conducting hypothesis testing, awareness of errors is crucial. A Type I error, in particular, occurs when we incorrectly reject a true null hypothesis. In other words, it is a false positive outcome—the conclusion that there’s an effect when there actually isn’t one.

In our exercise, the probability of making a Type I error corresponds to the significance level chosen for the test. Here, the significance level is 0.02 or 2%, suggesting a 2% risk of mistakenly rejecting a true null hypothesis. It's essential to weigh this risk carefully, as reducing it too much via a lower significance can increase the likelihood of a Type II error (failing to reject a false null hypothesis). Balance and understanding are key to effective hypothesis testing.
Understanding the P-value
The P-value is an essential part of hypothesis testing. It indicates the probability of observing sample data as extreme as the test statistic, under the assumption that the null hypothesis is true. In simpler terms, a smaller P-value suggests stronger evidence against the null hypothesis.

In our case, after calculating the Z-score, we determine the P-value. For a two-tailed test with \( Z = -1.6 \), we find the area in one tail of the distribution (0.0548) and double it, yielding a P-value of 0.1096.
This P-value is compared against the significance levels of 0.01 and 0.05. Since 0.1096 is greater than both, we do not reject the null hypothesis at these levels of significance. In essence, the P-value helps quantify the strength of evidence in hypothesis testing, leading to more informed decisions.

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Most popular questions from this chapter

The president of a university claims that the mean time spent partying by all students at this university is not more than 7 hours per week. A random sample of 40 students taken from this university showed that they spent an average of \(9.50\) hours partying the previous week with a standard deviation of \(2.3\) hours. Test at the \(2.5 \%\) significance level whether the president's claim is true. Explain your conclusion in words.

According to a 2008 survey by the Royal Society of Chemistry, \(30 \%\) of adults in Great Britain stated that they typically run the water for a period of 6 to 10 minutes while taking the shower (http://www.rsc.org/ AboutUs/News/PressReleases/2008/EuropeanShowerHabits.asp). Suppose that in a recent survey of 400 adults in Great Britain, 104 stated that they typically run the water for a period of 6 to 10 minutes when they take a shower. At the \(5 \%\) significance level, can you conclude that the proportion of all adults in Great Britain who typically run the water for a period of 6 to 10 minutes when they take a shower is less than 30 ? Use both the \(p\) -value and the critical value approaches.

A computer company that recently introduced a new software product claims that the mean time it takes to learn how to use this software is not more than 2 hours for people who are somewhat familiar with computers. A random sample of 12 such persons was selected. The following data give the times taken (in hours) by these persons to learn how to use this software. $$ \begin{array}{llllll} 1.75 & 2.25 & 2.40 & 1.90 & 1.50 & 2.75 \\ 2.15 & 2.25 & 1.80 & 2.20 & 3.25 & 2.60 \end{array} $$ Test at the \(1 \%\) significance level whether the company's claim is true. Assume that the times taken by all persons who are somewhat familiar with computers to learn how to use this software are approximately normally distributed.

According to the Magazine Publishers of America (www.magazine.org), the average visit at the magazines' Web sites during the fourth quarter of 2007 lasted \(4.145\) minutes. Forty-six randomly selected visits to magazine's Web sites during the fourth quarter of 2009 produced a sample mean visit of \(4.458\) minutes, with a standard deviation of \(1.14\) minutes. Using the \(10 \%\) significance level and the critical value approach, can you conclude that the length of an average visit to these Web sites during the fourth quarter of 2009 was longer than \(4.145\) minutes? Find the range for the \(p\) -value for this test. What will your conclusion be using this \(p\) -value range and \(\alpha=.10\) ?

In each of the following cases, do you think the sample size is large enough to use the normal distribution to make a test of hypothesis about the population proportion? Explain why or why not. a. \(n=30\) and \(p=.65\) b. \(n=70\) and \(p=.05\) c. \(n=60\) and \(p=.06\) d. \(n=900\) and \(p=.17\)

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