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91Ó°ÊÓ

(a) Find the relevant sample proportions in each group and the pooled proportion. (b) Complete the hypothesis test using the normal distribution and show all details. Test whether males are less likely than females to support a ballot initiative, if \(24 \%\) of a random sample of 50 males plan to vote yes on the initiative and \(32 \%\) of a random sample of 50 females plan to vote yes.

Short Answer

Expert verified
The sample proportions for males and females are 0.24 and 0.32 respectively, while the pooled proportion is 0.28. The hypothesis test is conducted under normal distribution, the calculation of which depends on the obtained test statistic and the set significance level. The conclusion of the test is based on whether the p-value is less than the significance level or not.

Step by step solution

01

Calculate Sample Proportions

Calculate the sample proportions for males and females. This is done by taking the percentage of those that plan to vote yes and dividing by 100. For males, it's \(24 \% = 0.24\) and for females, it's \(32 \% = 0.32\).
02

Calculate Pooled Proportion

Calculate the pooled proportion using the formula: \((x_1 + x_2) / (n_1 + n_2)\), where \(x_1\) and \(x_2\) are the number of 'yes' responses from males and females respectively, and \(n_1\) and \(n_2\) are the total number of males and females in the samples respectively. Here, \(x_1 = 0.24 * 50 = 12\), \(x_2 = 0.32 * 50 = 16\), \(n_1 = 50\), and \(n_2 = 50\). So the pooled proportion is \((12 + 16) / (50 + 50) = 0.28\).
03

Formulate Hypotheses

Formulate the null hypothesis and alternative hypothesis. The null hypothesis (\(H_0\)) is that there is no difference in the proportion of males and females supporting the ballot initiative, i.e., the proportion of males equals the proportion of females. The alternative hypothesis (\(H_a\)) is that the proportion of males supporting the ballot initiative is less than the proportion of females.
04

Perform Hypothesis Test

Use the normal distribution to perform the hypothesis test. The test statistic \(z\) is calculated using the formula: \[z = \frac {(p_1 - p_2) - 0} {\sqrt{(p(1 - p) (\frac{1}{n_1} + \frac{1}{n_2})}}\], where \(p_1\) and \(p_2\) are the proportions for males and females and \(p\) is the pooled proportion. Substituting the known values, the test statistic is calculated. Then, based on this test statistic and the significance level (usually 5%), the null hypothesis is either rejected or not rejected. Finally calculate the p-value.

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Key Concepts

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

Understanding Sample Proportion
Sample proportion is a key concept in statistics, especially when comparing two groups. It represents the ratio of success to the total number of trials in a sample. In the context of a survey, like the one in the exercise, it tells us the fraction of people who support a certain initiative out of the sampled population.
For instance, in our exercise, we calculate the sample proportion of males who plan to vote 'yes' on a ballot initiative. We know that 24% of 50 males plan to vote 'yes'. To find the sample proportion, we convert this percentage to a decimal by dividing by 100, resulting in 0.24. Similarly, for the 32% of females, converting to a decimal gives 0.32.
  • Sample proportion helps us understand the opinions within a specific group.
  • It's expressed as a decimal ranging from 0 to 1, representing full support in a survey.
  • In hypothesis testing, comparing sample proportions can reveal differences in group behavior.
What is Pooled Proportion?
Pooled proportion is utilized when conducting hypothesis tests that compare proportions from two groups.It combines the data from both groups to get an overall proportion that reflects both samples combined.
This overall proportion is useful because it provides a single estimate of the proportion based on all the data available, allowing for more nuanced statistical testing.
To calculate the pooled proportion, you combine the number of successes from both groups and divide by the total number of observations across both samples.In this case, using our exercise, we have 12 males and 16 females voting 'yes', out of a total of 100 individuals sampled (50 males + 50 females).Using the formula: \[ p = \frac {x_1 + x_2}{n_1 + n_2} \]We calculate the pooled proportion as:\[ \frac{12 + 16}{50 + 50} = 0.28 \]
  • Pooled proportion gives a cumulative success probability across groups.
  • It is crucial in calculating the test statistic for hypothesis testing.
  • This metric assumes that the probabilities of success in both groups are equal under the null hypothesis.
The Role of Normal Distribution in Hypothesis Testing
The normal distribution is an important tool in hypothesis testing.It allows statisticians to calculate the probability that a given outcome could occur, usually expressed as a z-score.This is especially helpful when we test claims like whether two sample proportions differ significantly.
In the exercise, we use the normal distribution to check if the difference between male and female support for the initiative is significant.
We assume that, under the null hypothesis, the distributions of support are the same, and thus their difference should follow a normal distribution. The test statistic calculated with pooled proportions follows a standard normal distribution, which is a special case with a mean of 0 and a standard deviation of 1.
By calculating the z-score using the formula:\[ z = \frac {(p_1 - p_2) - 0} {\sqrt{p(1 - p) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \]We can decide whether to reject the null hypothesis based on how extreme the z-score is.
  • Helps in determining the likelihood of a difference observed purely by chance.
  • Z-scores represent how many standard deviations an element is from the mean.
  • Used with significance levels (often 5%) to decide hypothesis acceptance or rejection.

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Most popular questions from this chapter

When we want \(95 \%\) confidence and use the conservative estimate of \(p=0.5,\) we can use the simple formula \(n=1 /(M E)^{2}\) to estimate the sample size needed for a given margin of error ME. In Exercises 6.40 to 6.43, use this formula to determine the sample size needed for the given margin of error. A margin of error of 0.05 .

Effect of Splitting the Bill Exercise 2.153 on page 105 describes a study to compare the cost of restaurant meals when people pay individually versus splitting the bill as a group. In the experiment half of the people were told they would each be responsible for individual meal costs and the other half were told the cost would be split equally among the six people at the table. The 24 people paying individually had a mean cost of 37.29 Israeli shekels with a standard deviation of 12.54 , while the 24 people splitting the bill had a higher mean cost of 50.92 Israeli shekels with a standard deviation of 14.33. The raw data can be found in SplitBill and both distributions are reasonably bell-shaped. Use this information to find and interpret a \(95 \%\) confidence interval for the difference in mean meal cost between these two situations.

Use a t-distribution and the given matched pair sample results to complete the test of the given hypotheses. Assume the results come from random samples, and if the sample sizes are small, assume the underlying distribution of the differences is relatively normal. Assume that differences are computed using \(d=x_{1}-x_{2}\). Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1}>\mu_{2}\) using the paired data in the following table: $$ \begin{array}{lllllllllll} \hline \text { Situation } & 1 & 125 & 156 & 132 & 175 & 153 & 148 & 180 & 135 & 168 & 157 \\ \text { Situation } & 2 & 120 & 145 & 142 & 150 & 160 & 148 & 160 & 142 & 162 & 150 \\ \hline \end{array} $$

A survey of 1000 adults in the US conducted in March 2011 asked "Do you favor or oppose 'sin taxes' on soda and junk food?" The proportion in favor of taxing these foods was \(32 \% .10\) (a) Find a \(95 \%\) confidence interval for the proportion of US adults favoring taxes on soda and junk food. (b) What is the margin of error? (c) If we want a margin of error of only \(1 \%\) (with \(95 \%\) confidence \()\), what sample size is needed?

Use a t-distribution and the given matched pair sample results to complete the test of the given hypotheses. Assume the results come from random samples, and if the sample sizes are small, assume the underlying distribution of the differences is relatively normal. Assume that differences are computed using \(d=x_{1}-x_{2}\). Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1} \neq \mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=15.7, s_{d}=12.2\) \(n_{d}=25 .\)

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