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Test \(H_{0}: p=0.2\) vs \(H_{a}: p \neq 0.2\) using the sample results \(\hat{p}=0.26\) with \(n=1000\)

Short Answer

Expert verified
The answer will depend on the calculated z-score and the subsequent p-value. Once these are calculated, they can be used to either accept or reject the null hypothesis.

Step by step solution

01

State the hypotheses

The null hypothesis \(H_0: p = 0.2\), this is the hypothesis we will test. The alternate hypothesis \(H_a: p \neq 0.2\), this means we are conducting a two-tailed test.
02

Calculate the z-score

The z-score can be calculated using the formula for z-score for proportion: \(z = \(\frac{\(\hat{p} - p\)}{\(\sqrt{\(\frac{p(1-p)}{n}\) }\)}\). Substituting the given values into the formula we get: \(z = \(\frac{0.26 - 0.2}{\sqrt{\(\frac{0.2(1-0.2)}{1000}\)}}\).
03

Find P-value

Using a z-table or standard normal distribution table, we can find the probability that a z-score is less than the calculated z-score for both tails. This is the p-value.
04

Compare P-value with the significance level

If the calculated p-value is less than the significance level (often 0.05), we reject the null hypothesis. If the p-value is more than the significance level, we do not reject the null hypothesis.
05

Draw the conclusion

Based on the p-value and the significance level, we make a decision about the hypothesis. If we rejected the null hypothesis, then there is evidence to support the alternative hypothesis, otherwise there is insufficient evidence to support the alternative hypothesis.

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Key Concepts

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

Z-Score Calculation
A Z-score is a statistical measurement that helps us understand how far away our sample proportion is from the population proportion under the null hypothesis. It's a standardized way to see where our sample compares to what's expected. Let's say you have a sample proportion \( \hat{p} = 0.26 \) and the population proportion under the null hypothesis is \( p = 0.2 \). First, recognize that you have 1000 observations, so \( n = 1000 \).

The Z-score formula for a sample proportion is:
  • \( z = \frac{(\hat{p} - p)}{\sqrt{\frac{p(1-p)}{n}}} \)
In this example, substitute \( \hat{p} = 0.26 \), \( p = 0.2 \), and \( n = 1000 \) into the formula. This simplifies to calculating how many standard deviations the observed sample proportion \( \hat{p} \) (0.26) is from the hypothesized population proportion \( p \) (0.2). The Z-score informs us if the difference is due to random chance or if it suggests an actual deviation from the hypothesized rate.

After substituting the values, you get:
  • \( z = \frac{(0.26 - 0.2)}{\sqrt{\frac{0.2(1-0.2)}{1000}}} \)
  • This will result in a numerical Z-score that needs further interpretation.
P-Value Interpretation
The p-value is a crucial component of hypothesis testing. It's the probability of obtaining a test statistic as extreme as the one observed, assuming that the null hypothesis is true. In simpler terms, it tells you how likely it is to get your observed results (or more extreme) if the null hypothesis was correct.

Once you've calculated the Z-score, you look up this value in a Z-table or use a calculator that provides the tail probabilities. The two-tailed test influences the p-value because it accounts for both tails of the distribution. This means you're considering extremes in both directions \((< \hat{p} \) and \( > \hat{p} \)).

To find the p-value for a two-tailed test:
  • Determine the probability of the Z-score in each tail.
  • Multiply by 2 because you consider both tails (left tail and right tail).
If your p-value is less than the predefined significance level (commonly 0.05), you reject the null hypothesis, suggesting that the observed sample provides enough evidence to state that the actual population proportion differs from the hypothesized proportion. If not, you fail to reject the null hypothesis.
Two-Tailed Test
A two-tailed test in hypothesis testing is used when there is potential for deviation in two directions. This is different from a one-tailed test where you're only interested in deviation in one direction. Here, you test if a sample proportion is either significantly greater than or significantly less than a specified population proportion.

In the context of this example, the null hypothesis \(H_0: p = 0.2\) is tested against the alternative hypothesis \(H_a: p eq 0.2\). This indicates that you're concerned with deviations on both the higher and lower sides of the population proportion. You want to know if the sample proportion \( \hat{p} = 0.26 \) deviates significantly in either direction from 0.2.

The two-tailed test requires evaluating both sides of the distribution curve. This involves determining the significance of deviations on either side of the hypothesized proportion by using distribution tables or specific software to calculate the corresponding p-values. Those p-values are then used to help you draw conclusions about the hypotheses. If your p-value from the two-tailed test falls below your chosen level of significance, then you reject the null hypothesis, showing evidence for a true difference between the sample and the population proportions.

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Most popular questions from this chapter

Table 6.29 gives a sample of grades on the first two quizzes in an introductory statistics course. We are interested in testing whether the mean grade on the second quiz is significantly higher than the mean grade on the first quiz. (a) Complete the test if we assume that the grades from the first quiz come from a random sample of 10 students in the course and the grades on the second quiz come from a different separate random sample of 10 students in the course. Clearly state the conclusion. (b) Now conduct the test if we assume that the grades recorded for the first quiz and the second quiz are from the same 10 students in the same order. (So the first student got a 72 on the first quiz and a 78 on the second quiz.) (c) Why are the results so different? Which is a better way to collect the data to answer the question of whether grades are higher on the second quiz? $$ \begin{array}{llllllllll} \hline \text { First Quiz } & 72 & 95 & 56 & 87 & 80 & 98 & 74 & 85 & 77 & 62 \\\ \text { Second Quiz } & 78 & 96 & 72 & 89 & 80 & 95 & 86 & 87 & 82 & 75 \\ \hline \end{array} $$

Standard Error from a Formula and Simulation In Exercises 6.15 to \(6.18,\) find the mean and standard error of the sample proportions two ways: (a) Use StatKey or other technology to simulate at least 1000 sample proportions. Give the mean and standard error and comment on whether the distribution appears to be normal. (b) Use the formulas in the Central Limit Theorem to compute the mean and standard error. Are the results similar to those found in part (a)? Sample proportions of sample size \(n=10\) from a population with \(p=0.2\)

IQ tests scale the scores so that the mean IQ score is \(\mu=100\) and standard deviation is \(\sigma=15\). Suppose that 30 fourth graders in one class are given such an IQ test that is appropriate for their grade level. If the students are really a random sample of all fourth graders, what is the chance that the average IQ score for the class is above \(105 ?\)

The AllCountries dataset shows that the percent of the population living in rural areas is \(57.3 \%\) in Egypt and \(21.6 \%\) in Jordan. Suppose we take random samples of size 400 people from each country, and compute the difference in sample proportions \(\hat{p}_{E}-\hat{p}_{J}\) where \(\hat{p}_{E}\) represents the sample proportion living in rural areas in Egypt and \(\hat{p}_{J}\) represents the sample proportion living in rural areas in Jordan. (a) Find the mean and standard deviation of the distribution of differences in sample proportions, \(\hat{p}_{E}-\hat{p}_{J}\) (b) If the sample sizes are large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. (c) Using the graph drawn in part (b), are we likely to see a difference in sample proportions as large in magnitude as \(0.4 ?\) As large as \(0.5 ?\) Explain.

In each case below, two sets of data are given for a two-tail difference in means test. In each case, which version gives a smaller \(\mathrm{p}\) -value relative to the other? (a) Both options have the same standard deviations and same sample sizes but: \(\begin{array}{lll}\text { Option 1 has: } & \bar{x}_{1}=25 & \bar{x}_{2}=23\end{array}\) Option 2 has: \(\quad \bar{x}_{1}=25 \quad \bar{x}_{2}=11\) (b) Both options have the same means \(\left(\bar{x}_{1}=22,\right.\) \(\left.\bar{x}_{2}=17\right)\) and same sample sizes but: Option 1 has: \(\quad s_{1}=15 \quad s_{2}=14\) $$ \text { Option 2 has: } \quad s_{1}=3 \quad s_{2}=4 $$ (c) Both options have the same means \(\left(\bar{x}_{1}=22,\right.\) \(\bar{x}_{2}=17\) ) and same standard deviations but: Option 1 has: \(\quad n_{1}=800 \quad n_{2}=1000\) Option 2 has: \(\quad n_{1}=25 \quad n_{2}=30\)

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