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Find the \(p\) -value for each of the following hypothesis tests. a. \(H_{0}: \mu=46, \quad H_{1}: \mu \neq 46, \quad n=40, \quad \bar{x}=49.60, \quad \sigma=9.7\) b. \(H_{0}: \mu=26, \quad H_{1}: \mu<26, \quad n=33, \quad \bar{x}=24.30, \quad \sigma=4.3\) c. \(H_{0}: \mu=18, \quad H_{1}: \mu>18, \quad n=55, \quad \bar{x}=20.50, \quad \sigma=7.8\)

Short Answer

Expert verified
The \(p\)-value for these tasks must be found for each separately. First calculate the z-score and then use that to look up the p-value for the appropriate tail(s) given the alternative hypothesis.

Step by step solution

01

Calculate z-score

The z-score measures the standard deviation-away the sample mean \(\bar{x}\) is far from the population mean under the null hypothesis \(\mu_0\). The formula for the z-score is: \(Z = (\bar{x} - \mu_0) / (\sigma / \sqrt{n})\). For task a, the parameters are \(\mu=46\), \(n=40\), \(\bar{x}=49.60\), and \(\sigma=9.7\). Substituting these values into the formula, calculate the z-score.
02

Find the p-value

Now look up the calculated z-score on the z-table or use a z-score calculator to find the corresponding p-value. Since task a is two-tailed, the p-value is twice the result from the table lookup or calculator.
03

Repeat step 1 and 2 for task b and c

For task b, the parameters are \(\mu=26\), \(n=33\), \(\bar{x}=24.30\), and \(\sigma=4.3\). Note that the alternative hypothesis is \(\mu<26\), so after the calculation, find the p-value as the left tail probability from the table or calculator. For task c, use the parameters \(\mu=18\), \(n=55\), \(\bar{x}=20.50\), and \(\sigma=7.8\). The alternative hypothesis is \(\mu>18\), so find the p-value as the right tail probability from the z-distribution.

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Key Concepts

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

Understanding the P-value
The p-value is a crucial concept in hypothesis testing. It helps us determine the strength of the evidence against the null hypothesis. Imagine you are performing an experiment and you want to know if your results are significant, the p-value tells you this likelihood.
The p-value is the probability that the observed data - or something more extreme - would occur if the null hypothesis is true. A low p-value indicates that such results are unlikely to occur by random chance, suggesting that the null hypothesis might not be true. Generally:
  • A p-value less than 0.05 usually leads us to reject the null hypothesis.
  • A p-value higher than 0.05 suggests there is not enough evidence to reject the null hypothesis.
The smaller the p-value, the stronger the evidence against the null hypothesis. It's important to remember that the p-value does not tell us the probability that the null hypothesis is true or false.
Decoding the Z-score
The z-score is a measure of how many standard deviations an element is from the mean. In hypothesis testing, it tells us how far away a sample mean is from the population mean stated in the null hypothesis. The formula for calculating a z-score is:
\[ Z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} \]
Here, \(\bar{x}\) is the sample mean, \(\mu_0\) is the population mean stated in the null hypothesis, \(\sigma\) is the standard deviation, and \(n\) is the sample size. A positive z-score indicates that the sample mean is higher than the population mean, while a negative z-score suggests it's lower. Heart of the z-score analysis is its ability to convert data into a standardized form, allowing us to look up probabilities related to the data using statistical tables.
The Null Hypothesis
The null hypothesis, often symbolized as \(H_0\), plays a central role in hypothesis testing. It is a statement that there is no effect or no difference, serving as the base assumption for statistical testing. For instance, your null hypothesis might claim that a particular educational program has no effect on students' test scores.
In hypothesis testing, you start by assuming the null hypothesis is true, and then examine the data to determine whether there is enough evidence to reject it. The beauty of the null hypothesis is its simplicity; it keeps our expectations neutral until we can prove otherwise. Only when the evidence is strong enough — typically when the p-value is low — do we reject the null hypothesis.
Exploring the Alternative Hypothesis
The alternative hypothesis is what you might consider the research hypothesis. It's the statement that suggests there is a significant effect or a difference. Symbolized as \(H_1\), the alternative hypothesis is what the researcher aims to support.
Depending on the nature of your test, the alternative hypothesis can be:
  • Two-tailed: Suggests the mean is not equal to a specific value.
  • Left-tailed: Suggests the mean is less than a specific value.
  • Right-tailed: Suggests the mean is greater than a specific value.
These indicate the direction of the effect being tested. For example, in a study to examine if a drug lowers blood pressure, the alternative hypothesis might suggest that blood pressure is lower after taking the drug compared to before. Determining whether to reject the null hypothesis in favor of the alternative hypothesis depends on the strength of the evidence provided by the data analysis.

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Most popular questions from this chapter

Write the null and alternative hypotheses for each of the following examples. Determine if each is a case of a two-tailed, a left-tailed, or a right-tailed test. a. To test if the mean amount of time spent per week watching sports on television by all adult men is different from \(9.5\) hours b. To test if the mean amount of money spent by all customers at a supermarket is less than \(\$ 105\) c. To test whether the mean starting salary of college graduates is higher than \(\$ 47,000\) per year d. To test if the mean waiting time at the drive-through window at a fast food restaurant during rush hour differs from 10 minutes e. To test if the mean hours spent per week on house chores by all housewives is less than 30

food company is planning to market a new type of frozen yogurt. However, before marketing thit yogurt, the company wants to find what percentage of the people like it. The company's management has decided that it will market this yogurt only if at least \(35 \%\) of the people like it. The company's researcl department selected a random sample of 400 persons and asked th taste this yogurt. Of these 400 persons, 112 said they liked a. Testing at a \(2.5 \%\) significance level, can you conclude that the company should market this yogurt b. What will your decision be in part a if the probability of making a Type I error is zero? Explain Make the test of part a using the \(p\) -value approach.min

Consider \(H_{0}=p=.70\) versus \(H_{1}: p \neq .70\). a. A random sample of 600 observations produced a sample proportion equal to .68. Using \(\alpha=.01\), would you reject the null hypothesis? b. Another random sample of 600 observations taken from the same population produced a sample proportion equal to. \(76 .\) Using \(\alpha=.01\), would you reject the null hypothesis? Comment on the results of parts a and \(\mathrm{b}\).

Make the following tests of hypotheses. a. \(H_{0}: \mu=80, \quad H_{1}: \mu \neq 80, \quad n=33, \quad \bar{x}=76.5, \quad \sigma=15, \quad \alpha=.10\) b. \(H_{0}=\mu=32, \quad H_{1}: \mu<32, \quad n=75, \quad \bar{x}=26.5, \quad \sigma=7.4, \quad \alpha=.01\) c. \(H_{0}=\mu=55, \quad H_{1}: \mu>55, \quad n=40, \bar{x}=60.5, \quad \sigma=4, \quad \alpha=.05\)

Consider the following null and alternative hypotheses: $$ H_{0}=\mu=60 \text { versus } H_{1}: \mu>60 $$ Suppose you perform this test at \(\alpha=.01\) and fail to reject the null hypothesis. Would you state that the difference between the hypothesized value of the population mean and the observed value of the sample mean is "statistically significant" or would you state that this difference is "statistically not significant"? Explain.

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