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91Ó°ÊÓ

As reported on carefair.com (November 15,2006\(), 40 \%\) of women aged 30 years and older would rather get Botox injections than spend a week in Paris. In a recent survey of 400 women aged 65 years and older, 108 women would rather get Botox injections than spend a week in Paris. Using a \(10 \%\) significance level, perform a test of hypothesis to determine whether the current percentage of women aged 65 years or older who would rather get Botox injections than spend a week in Paris is less than \(40 \%\). Use both the \(p\) -value and the critical-value approaches.

Short Answer

Expert verified
According to the critical value approach, if the test statistic is less than the critical value of -1.28, we would reject the null hypothesis and conclude the proportion of women aged 65 and older preferring botox is less than 40%. Similarly with the p-value approach, we would also reject the null hypothesis if the calculated p-value is less than the significance level of 10%. The exact numerical result will depend on the calculated test statistic and p-value using statistical tools/software.

Step by step solution

01

Formulate the Hypotheses

The null hypothesis assumes that there is no change, which in this case would be that the proportion of women aged 65 or older preferring botox injections over a week in Paris is equal to 40 percent or \( p = 0.40 \). Conversely, the alternative hypothesis is that the proportion is less than 40 percent or \( p < 0.40 \).
02

Gather Data

From the survey of 400 women aged 65 and older, it's mentioned that 108 women prefer getting Botox injections. Thus, we can calculate the sample proportion as \( p = 108/400 = 0.27 \).
03

Calculate the Test Statistic

The test statistic for a proportion is calculated using the Z score formula, which is \( Z = (p - \mu_{p}) / \sqrt{(p * (1 - p) / n)} \), where \( p \) is the sample proportion, \( \mu_{p} \) is the population proportion, and \( n \) is the sample size. Substituting the provided values, we get \( Z = (0.27 - 0.40) / \sqrt{(0.40 * (1 - 0.40) / 400)} \).
04

Calculate the p-value and Define Critical Value

The p-value can be found using the standard normal distribution table or a calculator that can perform this function with the test statistic as the input. The critical value for a one-tailed test at the 10% significance level is -1.28. This is chosen based on the condition laid out in the null and alternative hypotheses.
05

Determine the Conclusion

Compare the test statistic with the critical value and p-value with the significance level. If the test statistic is less than the critical value and the p-value is less than the significance level, the null hypothesis is rejected in favor of the alternative hypothesis. Otherwise, there is not enough evidence 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.

Significance Level
Significance level is a crucial concept in hypothesis testing. It represents the probability of rejecting the null hypothesis when it is actually true. In simpler terms, it tells us how much error we are willing to accept when making a decision using statistical evidence.

For example, in our exercise, the significance level is set at 10%, or 0.10. This means that there is a 10% risk of concluding that the proportion of women preferring Botox over a trip to Paris has changed, even if it has not.

Choosing the right significance level involves balancing the risks of Type I (false positive) and Type II (false negative) errors. A lower significance level reduces the risk of Type I errors but increases the risk of Type II errors, and vice versa.
Test Statistic
The test statistic is a standardized value that is calculated from sample data during a hypothesis test. It is used to determine whether to reject the null hypothesis.

In the exercise, we calculate the test statistic using the Z score formula. The formula for the test statistic when testing proportions is:

\[Z = \frac{(p - \mu_{p})}{\sqrt{\left(\frac{p(1 - p)}{n}\right)}}\]
where:
  • \(p\) is the sample proportion (0.27),
  • \(\mu_{p}\) is the population proportion (0.40),
  • \(n\) is the sample size (400).

The test statistic helps us understand how far the sample proportion is from the population proportion under the null hypothesis, measured in standard deviations.
p-value
The p-value is a measure that helps determine the strength of evidence against the null hypothesis. It tells us the probability of obtaining a sample statistic as extreme as the one observed, assuming the null hypothesis is true.

If the p-value is smaller than the significance level, it suggests that the observed data is unlikely under the null hypothesis, and thus, the null hypothesis can be rejected. In our exercise, we calculated the test statistic and found the p-value accordingly.

If the p-value is greater than 0.10 (our significance level), the evidence is not strong enough to reject the null hypothesis.

Remember, a lower p-value reflects stronger evidence against the null hypothesis.
Critical Value
The critical value is a threshold that helps decide whether to reject the null hypothesis in a hypothesis test. It is based on the significance level and the specific distribution of the test statistic.

For a one-tailed test at the 10% significance level, as in this exercise, the critical value is -1.28. We compare the test statistic to this critical value.

If the test statistic is less than the critical value, it indicates enough evidence to reject the null hypothesis in favor of the alternative hypothesis.

Critical values depend on the context of the test, including whether it’s one-tailed or two-tailed, and the chosen significance level.
Null and Alternative Hypotheses
In hypothesis testing, the null hypothesis ( \(H_0\)) represents a statement of no effect or no difference, while the alternative hypothesis (\(H_1\)) represents what we aim to support.

In this context, the null hypothesis states that 40% of women aged 65 or older prefer Botox over a week in Paris ( \(p = 0.40\)). Conversely, the alternative hypothesis suggests that less than 40% prefer Botox ( \(p < 0.40\)).

Formulating these hypotheses is the first crucial step in any hypothesis testing. It gives a clear direction for the test and determines how we interpret the results from comparing test statistics with critical values or examining p-values.

Understanding these hypotheses is vital because they guide all subsequent steps, from data collection to deciding on the test conclusion.

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Most popular questions from this chapter

According to an article on PCMag.com, Facebook users spend an average of 190 minutes per month checking and updating their Facebook page (Source: http://www.pcmag.com/article2/ \(0,2817,2342757,00\) asp). A random sample of 55 college students aged 18 to 22 years with Facebook accounts resulted in a sample mean time and sample standard deviation of \(219.50\) and \(69.30\) minutes per month, respectively. a. Using \(\alpha=.025\), can you conclude that the average time spent per month checking and updating their Facebook pages by all college students aged 18 to 22 years who have Facebook accounts is higher than 190 minutes? Use the critical value approach. b. Find the range of the \(p\) -value for the test of part a. What is your conclusion with \(\alpha=.025\) ?

Consider the following null and alternative hypotheses: $$ H_{0}: p=.44 \text { versus } H_{1}: p<.44 $$ A random sample of 450 observations taken from this population produced a sample proportion of \(.39 .\) 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 \(\alpha=.025\) ?

You read an article that states " 50 hypothesis tests of \(H_{0}: \mu=35\) versus \(H_{1}: \mu \neq 35\) were performed using \(\alpha=.05\) on 50 different samples taken from the same population with a mean of \(35 .\) Of these, 47 tests failed to reject the null hypothesis." Explain why this type of result is not surprising.

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 \(\alpha=05\) ?

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.

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