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Chapter 9: Problem 77

Goin' to the Chapel? If you choose to marry, what type of wedding site will you pick? A USA Today snapshot claims that \(43 \%\) of all brides-to-be choose a house of worship for their wedding site. \({ }^{24}\) In a follow-up study, a random sample of 100 bridesto-be found that 46 of those sampled had chosen or would choose a house of worship for their wedding site. Does this sample contradict the reported \(43 \%\) figure? Test at the \(\alpha=.05\) level of significance.

Short Answer

Expert verified
Answer: No, there is no sufficient evidence to conclude that the proportion of brides choosing a house of worship for their wedding site is different from the reported 43%.

Step by step solution

01

State the null and alternative hypothesis

The null hypothesis (H鈧�) is that the proportion of brides choosing a house of worship for their wedding site is 43%, as reported by USA Today. The alternative hypothesis (H鈧�) is that the proportion is not equal to 43%, meaning our sample contradicts the reported figure. In mathematical terms, we have: H鈧�: p = 0.43 H鈧�: p 鈮� 0.43
02

Determine the test statistic and critical values

To perform a hypothesis test for a proportion, we use the z-test. We can calculate the z-value using the formula: z = (p虃 - p鈧�) / 鈭�(p鈧�(1 - p鈧�) / n) where p虃 is the sample proportion, p鈧� is the population proportion, and n is the sample size. With a significance level (伪) of 0.05 and a two-tailed test (since the alternative hypothesis uses 鈮�), we can find the critical values using a standard normal distribution table. The critical values are -1.96 and 1.96.
03

Calculate the sample proportion and test statistic

First, we calculate the sample proportion (p虃) by dividing the number of brides choosing a house of worship (46) by the sample size (100): p虃 = 46/100 = 0.46 Next, we calculate the test statistic (z) using the formula: z = (0.46 - 0.43) / 鈭�(0.43(1 - 0.43) / 100) z = 0.03 / 鈭�(0.43(0.57) / 100) z = 0.03 / 0.0495 鈮� 0.61
04

Compare test statistic with critical values

We now compare the test statistic (0.61) with the critical values (-1.96 and 1.96). Since -1.96 < 0.61 < 1.96, we fail to reject the null hypothesis at the 0.05 significance level.
05

Draw the conclusion

Since we fail to reject the null hypothesis, we cannot say that the sample contradicts the reported 43% figure. There is no sufficient evidence to conclude that the proportion of brides choosing a house of worship for their wedding site is different from the reported 43%.

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Key Concepts

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

z-test
The z-test is a statistical method used to determine if there's a significant difference between a sample mean and a population mean. When we talk about a z-test for proportions, we're comparing the proportion from our sample to a known population proportion.
Here, we're testing if the proportion of brides using a house of worship is different from 43%.
  • Formula: The z-value can be calculated using \[ z = \frac{{\hat{p} - p_0}}{\sqrt{\frac{{p_0(1 - p_0)}}{n}}} \]where \( \hat{p} \) is the sample proportion, \( p_0 \) is the population proportion, and \( n \) is the sample size.
  • Usage: The z-test is helpful when the sample size is large enough, typically \( n \geq 30 \).
  • Outcome: It provides a test statistic (z-score) that can be compared with critical values to determine the result concerning the null hypothesis.
Applying it allows us to see if our sample data provides enough evidence to challenge existing population claims.
null hypothesis
The null hypothesis, often denoted by \( H_0 \), serves as a starting point for statistical hypothesis testing.
It proposes that there is no effect or no difference between the population parameter and the sample statistic.
For this exercise, the null hypothesis states that the population proportion of brides choosing a house of worship is exactly 43%.
  • Purpose: The purpose of the null hypothesis is to test the validity of a claim using sample data.
  • Assumption: We assume that any observed difference is due to random sampling variability.
  • Notation: It is symbolically represented as \( H_0: p = 0.43 \) in this scenario.
By not rejecting the null hypothesis in this case, we conclude there is no significant contradiction to the reported proportion.
alternative hypothesis
Contrasting the null hypothesis, the alternative hypothesis, represented by \( H_1 \), suggests that there is an effect or a difference.
In our context, the alternative hypothesis posits that the proportion of brides choosing a house of worship is not equal to 43%.
  • Formulation: It is formulated to challenge the null hypothesis. For this exercise, we have \( H_1: p eq 0.43 \).
  • Direction: The alternative hypothesis can be one-sided or two-sided. Our example is two-sided because we're testing for a difference, not a specific direction of difference.
  • Implications: If evidence sufficiently supports \( H_1 \), it suggests that our sample contradicts the null hypothesis.
In this particular exercise, the result did not support \( H_1 \), leaving the null hypothesis unchallenged.
significance level
The significance level, denoted \( \alpha \), is a crucial concept in hypothesis testing.
It determines the threshold for deciding whether or not to reject the null hypothesis.
In this exercise, we used a significance level of 0.05, which means we're willing to accept a 5% chance of incorrectly rejecting the null hypothesis.
  • Definition: The significance level is the probability of rejecting the null hypothesis when it is actually true.
  • Common Levels: Typical values for \( \alpha \) are 0.05, 0.01, and 0.10, depending on the field of study and the precision needed.
  • Decision Rule: If the test statistic falls beyond the critical values defined by \( \alpha \), we reject \( H_0 \). Otherwise, we fail to reject it.
This level helps scientists balance the risk of making a type I error, providing confidence in the test conclusion.

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Most popular questions from this chapter

Find the \(p\) -value for the following large-sample \(z\) tests: a. A right-tailed test with observed \(z=1.15\) b. A two-tailed test with observed \(z=-2.78\) c. A left-tailed test with observed \(z=-1.81\)

Invasive Species In a study of the pernicious giant hogweed, Jan Pergl \(^{1}\) and associates compared the density of these plants in two different sites within the Caucasus region of Russia. In its native area, the average density was found to be 5 plants/m \(^{2}\). In an invaded area in the Czech Republic, a sample of \(n=50\) plants produced an average density of 11.17 plants/m \(^{2}\) with a standard deviation of 3.9 plants \(/ \mathrm{m}^{2}\) a. Does the invaded area in the Czech Republic have an average density of giant hogweed that is different from \(\mu=5\) at the \(\alpha=.05\) level of significance? b. What is the \(p\) -value associated with the test in part a? Can you reject \(H_{0}\) at the \(5 \%\) level of significance using the \(p\) -value?

Find the appropriate rejection regions for the large-sample test statistic \(z\) in these cases: a. A right-tailed test with \(\alpha=.01\) b. A two-tailed test at the \(5 \%\) significance level

Suppose you wish to detect a difference between \(\mu_{1}\) and \(\mu_{2}\) (either \(\mu_{1}>\mu_{2}\) or \(\mu_{1}<\mu_{2}\) ) and, instead of running a two-tailed test using \(\alpha=.05,\) you use the following test procedure. You wait until you have collected the sample data and have calculated \(\bar{x}_{1}\) and \(\bar{x}_{2}\). If \(\bar{x}_{1}\) is larger than \(\bar{x}_{2},\) you choose the alternative hypothesis \(H_{\mathrm{a}}: \mu_{1}>\mu_{2}\) and run a one-tailed test placing \(\alpha_{1}=.05\) in the upper tail of the \(z\) distribution. If, on the other hand, \(\bar{x}_{2}\) is larger than \(\bar{x}_{1},\) you reverse the procedure and run a one-tailed test, placing \(\alpha_{2}=.05\) in the lower tail of the \(z\) distribution. If you use this procedure and if \(\mu_{1}\) actually equals \(\mu_{2},\) what is the probability \(\alpha\) that you will conclude that \(\mu_{1}\) is not equal to \(\mu_{2}\) (i.e., what is the probability \(\alpha\) that you will incorrectly reject \(H_{0}\) when \(H_{0}\) is true)? This exercise demonstrates why statistical tests should be formulated prior to observing the data.

A random sample of \(n=1400\) observations from a binomial population produced \(x=529 .\) a. If your research hypothesis is that \(p\) differs from .4, what hypotheses should you test? b. Calculate the test statistic and its \(p\) -value. Use the \(p\) -value to evaluate the statistical significance of the results at the \(1 \%\) level. c. Do the data provide sufficient evidence to indicate that \(p\) is different from \(.4 ?\)
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