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Calculate the \(p\) -value, given \(H_{a}: \mu \neq 245\) and \(z \star=1.1\)

Short Answer

Expert verified
The p-value for this two-tailed test with given z-score of 1.1 is approximately 0.2714.

Step by step solution

01

Identify Type of Test

Because the alternative hypothesis (\(H_{a}\)) implies a value different from 245 (either greater or less), it indicates a two-tailed test.
02

Locate Z-Score on Standard Normal Table

Using a standard normal distribution table, or calculator with statistical functions, find the area that corresponds to the given z-score of 1.1. This area is also known as the cumulative probability value, which can be denoted as \(P(Z \leq 1.1)\). The result from the table is approximately 0.8643, thus \(P(Z \leq 1.1) = 0.8643\).
03

Calculate One-tail P-value

Subtract the cumulative probability value obtained from Step 2 from 1 to get the one-tail p-value: \(P(Z > 1.1) = 1- P(Z \leq 1.1) = 1 - 0.8643 = 0.1357\). This expresses the probability that a z-score is greater than 1.1 under the standard normal curve.
04

Calculate Two-tail P-value

Since this is a two-tailed test, the total p-value will be twice the one-tail p-value obtained in Step 3. So, \(p = 2 * P(Z > 1.1) = 2 * 0.1357 = 0.2714\). The result is the p-value for this two-tailed hypothesis test.

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Key Concepts

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

Hypothesis Testing
Understanding hypothesis testing is crucial for anyone delving into the realm of statistics. At its core, hypothesis testing is a method used to decide whether there is enough evidence to reject a null hypothesis, represented as \( H_0 \). The null hypothesis typically suggests that there is no effect or no difference, and it serves as a default or starting assumption to be challenged.The process of hypothesis testing involves proposing an alternative hypothesis, \( H_a \), which is a statement we are seeking evidence for, often suggesting that there is an effect or a difference. In the given exercise, we see an example where \( H_a: \mu eq 245 \) signifies that the mean is not equal to 245.To conduct a hypothesis test, one calculates a test statistic, which is then compared against a critical value or converted into a p-value for making a decision. If the p-value is less than a predetermined significance level, often \( \alpha = 0.05 \), we reject the null hypothesis. Conversely, if the p-value is larger, we fail to reject the null hypothesis. This procedure offers a systematic way to make decisions that account for uncertainty and variability in data.
Z-Score
A z-score provides a way of describing how far and in what direction, a data point is from the mean, measured in terms of standard deviations. It is a standardized score that helps to compare results from different tests or analyze individual scores within the same test.The formula to compute the z-score of a data point is:\[ z = \frac{(X - \mu)}{\sigma} \]where \( X \) represents the data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. In our problem, we're given a z-score of 1.1, which means that the data point in question lies 1.1 standard deviations above the mean of the standard normal distribution.
Standard Normal Distribution
The standard normal distribution, a special case of the normal distribution, is central to statistics. It has a mean of 0 and a standard deviation of 1. When a dataset follows a normal distribution, converting the data to z-scores translates the data into the standard normal distribution.Useful properties of the standard normal distribution include its symmetry and the fact that percentages of data falling within a certain number of standard deviations from the mean are fixed. For instance, about 68% of the data falls within 1 standard deviation from the mean, and approximately 95% falls within 2 standard deviations.In hypothesis testing, we often use the standard normal distribution to find p-values, as it allows us to determine the probability of observing a test statistic as or more extreme than the one calculated.
Two-Tailed Test
A two-tailed test is a statistical test where the area of interest is in both tails of the distribution. This test is used when the alternative hypothesis specifies a value that is different from the null hypothesis mean, meaning it could be either greater than or less than that mean but not equal to it.In such tests, we are looking for evidence against the null hypothesis in either direction. Accordingly, we split our significance level between both tails of the distribution. If \( z_* \) is the calculated test statistic from our data, to find the p-value, we determine the probability of seeing a z-score as extreme as \( z_* \) or more so in both directions, which involves doubling the p-value obtained from one tail, as seen in the exercise.Two-tailed tests are the appropriate choice when we don't have a specific direction of interest beforehand, as they provide a more conservative estimate compared to a one-tailed test by accounting for extremities in the data in either direction.

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Most popular questions from this chapter

Do you drink the recommended amount of water each day? Most Americans don't! On average, Americans drink 4.6 eight-oz servings of water a day. A sample of 42 education professionals was randomly selected and their water consumption for a 24 -hour period was monitored; the mean amount consumed was 39.3 oz. Assuming the amount of water consumed daily by adults is normally distributed and the standard deviation is 11.2 oz, is there sufficient evidence to show that education professionals consume, on average, more water daily than the national average? Use \(\alpha=0.05\)

The weights of full boxes of a certain kind of cereal are normally distributed with a standard deviation of 0.27 oz. A sample of 18 randomly selected boxes produced a mean weight of 9.87 oz. a. Find the \(95 \%\) confidence interval for the true mean weight of a box of this cereal. b. Find the \(99 \%\) confidence interval for the true mean weight of a box of this cereal. c. What effect did the increase in the level of confidence have on the width of the confidence interval?

A random sample of the scores of 100 applicants for clerk-typist positions at a large insurance company showed a mean score of \(72.6 .\) The preparer of the test maintained that qualified applicants should average \(75.0 .\) a. Determine the \(99 \%\) confidence interval for the mean score of all applicants at the insurance company. Assume that the standard deviation of test scores is 10.5 b. Can the insurance company conclude that it is getting qualified applicants (as measured by this test)?

The null hypothesis, \(H_{o}: \mu=250,\) was tested against the alternative hypothesis, \(H_{a}: \mu<250 .\) A sample of \(n=85\) resulted in a calculated test statistic of \(z \star=-1.18 .\) If \(\sigma=22.6,\) find the value of the sample mean, \(\bar{x}\). Find the sum of the sample data, \(\Sigma x\)

Assume that \(z\) is the test statistic and calculate the value of \(z \star\) for each of the following: a. \(\quad H_{o}: \mu=51, \sigma=4.5, n=40, \bar{x}=49.6\) b. \(\quad H_{o}: \mu=20, \sigma=4.3, n=75, \bar{x}=21.2\) c. \(\quad H_{o}: \mu=138.5, \sigma=3.7, n=14, \bar{x}=142.93\) d. \(\quad H_{o}: \mu=815, \sigma=43.3, n=60, \bar{x}=799.6\)

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