/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 Use the following information to... [FREE SOLUTION] | 91Ó°ÊÓ

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Use the following information to answer the next five exercises. Two metal alloys are being considered as material for ball bearings. The mean melting point of the two alloys is to be compared. 15 pieces of each metal are being tested. Both populations have normal distributions. The following table is the result. It is believed that Alloy Zeta has a different melting point. $$\begin{array}{|l|l|l|}\hline & {\text { Sample Mean Melting Temperatures }\left(^{\circ} F\right)} & {\text { Population Standard Deviation }} \\\ \hline \text { Alloy Gamma } & {800}&{95} \\ \hline \text { Alloy zeta } & {900} &{105} \\ \hline\end{array}$$ Draw the graph of the p-value.

Short Answer

Expert verified
The graph shows shaded tails at \( z = -2.52 \) and \( z = 2.52 \).

Step by step solution

01

State the Hypotheses

We need to compare the mean melting points of the two alloys. Our null hypothesis is that the mean melting points are equal: \( H_0: \mu_{Gamma} = \mu_{Zeta} \). The alternative hypothesis is that the mean melting points are not equal: \( H_a: \mu_{Gamma} eq \mu_{Zeta} \). This is a two-tailed test as we are testing for inequality.
02

Calculate the Test Statistic

Use the formula for the test statistic for two means with known population standard deviations:\[z = \frac{(\bar{x}_1 - \bar{x}_2) - 0}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\]Substitute the values: \( \bar{x}_1 = 800 \), \( \bar{x}_2 = 900 \), \( \sigma_1 = 95 \), \( \sigma_2 = 105 \), and \( n_1 = n_2 = 15 \):\[z = \frac{800 - 900}{\sqrt{\frac{95^2}{15} + \frac{105^2}{15}}} = \frac{-100}{\sqrt{\frac{9025}{15} + \frac{11025}{15}}}\]Simplify to get \( z \approx -2.52 \).
03

Find the P-value

Look up the calculated \( z \) value in the standard normal distribution table or use a calculator. Since this is a two-tailed test, we find the probability for \( z \approx -2.52 \) and \( z \approx 2.52 \). The p-value is \( P(Z < -2.52) + P(Z > 2.52) \).
04

Draw the Graph of the P-value

On the x-axis, mark the standard normal distribution. Shade the tails beyond \( z = -2.52 \) and \( z = 2.52 \) to represent the p-value areas. This shaded area is the p-value. The graph is symmetric with shaded regions in both tails.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a statistical method used to make decisions based on data. It's especially useful to determine whether a suspected relationship exists or if outcomes are due to random chance.
In our specific case of comparing two metal alloys, we have a null hypothesis (often denoted as \( H_0 \)) and an alternative hypothesis (\( H_a \)).
  • Null Hypothesis \( H_0 \): The mean melting points of the two alloys are equal (\( \mu_{Gamma} = \mu_{Zeta} \)).
  • Alternative Hypothesis \( H_a \): The mean melting points of the two alloys are not equal (\( \mu_{Gamma} eq \mu_{Zeta} \)).
After establishing these hypotheses, the goal is to use sample data to decide whether to reject the null hypothesis in favor of the alternative. If the evidence (data) is strong enough against \( H_0 \), we reject it. Otherwise, we do not reject \( H_0 \). Hypothesis testing involves choosing a significance level, usually denoted as \( \alpha \), which measures how extreme observed data must be to reject \( H_0 \). Common choices for \( \alpha \) are 0.05 or 0.01.
Standard Normal Distribution
The standard normal distribution is a continuous probability distribution, characterized by a bell-shaped curve that is perfectly symmetrical around the mean. It plays a crucial role in statistics, especially in hypothesis testing, as it provides a common framework for determining probabilities.
Key characteristics include:
  • The mean is 0.
  • The standard deviation is 1.
For hypothesis testing, we often use the z-distribution, which is a standard normal distribution applied via z-scores. A z-score measures how many standard deviations an element is from the mean. In our example, we calculate a z-score to understand the difference in mean melting points between two alloys. This z-score tells us where in the standard normal distribution our data point lies.
P-Value Calculation
The p-value helps us determine the significance of our results in hypothesis testing.
To calculate the p-value, we first determine the z-score, which, for our example, is approximately \(-2.52\). We then find the probability associated with this z-score from the standard normal distribution. Since we are conducting a two-tailed test, we need to consider both tails of the distribution.Specifically, the p-value is calculated as:
  • \( P(Z < -2.52) \)
  • \( P(Z > 2.52) \)
  • The p-value is the sum of these probabilities.
A small p-value (typically \( < 0.05 \)) indicates strong evidence against the null hypothesis, suggesting that we reject \( H_0 \). Conversely, a large p-value suggests we fail to reject \( H_0 \). The p-value provides a measure of the evidence against the null hypothesis.
Two-Tailed Test
A two-tailed test is used when we are interested in deviations in both directions from the null hypothesis. Unlike one-tailed tests, which test for an effect in a single direction, two-tailed tests check for any significant difference outside the specified range.
In our alloy example, the two-tailed test means we are interested in whether the mean melting points are either higher or lower for the alloys – not just one direction.
What this involves:
  • Checking for significant results at both ends of the distribution.
  • Calculating the p-value for both tails, then summing these probabilities.
Graphically, it means shading both tails of the standard distribution curve beyond the calculated z-scores of \(-2.52\) and \(2.52\). If these shaded areas represent a small total probability (less than the significance level), it leads us to reject the null hypothesis, supporting the claim of a difference in means.

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Most popular questions from this chapter

Use the following information to answer the next 15 exercises: Indicate if the hypothesis test is for a. independent group means, population standard deviations, and/or variances known b. independent group means, population standard deviations, and/or variances unknown c. matched or paired samples d. single mean e. two proportions f. single proportion A new WiFi range booster is being offered to consumers. A researcher tests the native range of 12 different routers under the same conditions. The ranges are recorded. Then the researcher uses the new WiFi range booster and records the new ranges. Does the new WiFi range booster do a better job?

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Use the following information to answer the next 12 exercises: The U.S. Center for Disease Control reports that the mean life expectancy was 47.6 years for whites born in 1900 and 33.0 years for nonwhites. Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124 whites, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 nonwhites, the mean life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test to see if the mean life spans in the county were the same for whites and nonwhites. At a pre-conceived \(\alpha=0.05,\) what is your: a. Decision: b. Reason for the decision: c. Conclusion (write out in a complete sentence):

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