/*! 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 40 Use the following information to... [FREE SOLUTION] | 91Ó°ÊÓ

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Use the following information to answer the next five exercises. A researcher is testing the effects of plant food on plant growth. Nine plants have been given the plant food. Another nine plants have not been given the plant food. The heights of the plants are recorded after eight weeks. The populations have normal distributions. The following table is the result. The researcher thinks the food makes the plants grow taller. $$\begin{array}{|l|l|l|}\hline \text { Plant Group } & {\text { Sample Mean Height of Plants (inches) }} & {\text { Population Standard Deviation }} \\\ \hline \text { Food } & {16} & {2.5} \\ \hline \text { No food } & {14} & {1.5} \\ \hline\end{array}$$ Draw the graph of the p-value.

Short Answer

Expert verified
The p-value area is 0.020, indicating significant results.

Step by step solution

01

Define the Hypotheses

First, we need to establish the null hypothesis and the alternative hypothesis. Let \( \mu_F \) represent the mean height of plants with food and \( \mu_N \) the mean height without food. The null hypothesis (\( H_0 \)) states that the plant food has no effect, so \( \mu_F = \mu_N \). The alternative hypothesis (\( H_1 \)) suggests the plant food increases height, so \( \mu_F > \mu_N \).
02

Determine the Test Statistic

Since we are comparing the means of two independent groups with known standard deviations, we'll use the z-test for comparison. The z-test statistic is calculated as follows: \[ z = \frac{(\bar{x}_F - \bar{x}_N) - 0}{\sqrt{\frac{\sigma_F^2}{n_F} + \frac{\sigma_N^2}{n_N}}} \]Substitute \( \bar{x}_F = 16 \), \( \bar{x}_N = 14 \), \( \sigma_F = 2.5 \), \( \sigma_N = 1.5 \), and \( n_F = n_N = 9 \).
03

Calculate the Z-score

Calculate the z-score using the formula:\[ z = \frac{16 - 14}{\sqrt{\frac{2.5^2}{9} + \frac{1.5^2}{9}}} \]First, calculate the denominator:\[ \sqrt{\frac{6.25}{9} + \frac{2.25}{9}} = \sqrt{\frac{8.5}{9}} = \sqrt{0.9444} \approx 0.9718 \]Then calculate the z-score:\[ z = \frac{2}{0.9718} \approx 2.057 \]
04

Find the P-value

The p-value represents the probability of observing a z-score as extreme as 2.057 or more in a standard normal distribution, assuming the null hypothesis is true. Use the standard normal table to find that \( P(Z > 2.057) \). This p-value can also be found using statistical software or a calculator. The approximate p-value is 0.020.
05

Draw the Graph

On a standard normal distribution graph, plot the critical region associated with \( Z = 2.057 \). Shade the region to the right of \( z = 2.057 \), which represents the p-value. The area of this shaded region is the p-value, which is 0.020.

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Key Concepts

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

Z-test
The Z-test is a statistical procedure used when the distribution of data is normal, and the sample size is relatively large, generally considered to be over 30. However, when population standard deviations are known, the Z-test can be used for smaller samples as well, as in the plant growth study. The Z-test helps determine whether there is a significant difference between the means of two groups. In this example, we compare the heights of plants treated with food versus those not given food.

The Z-test formula is given by:
  • \[ z = \frac{(\bar{x}_F - \bar{x}_N) - 0}{\sqrt{\frac{\sigma_F^2}{n_F} + \frac{\sigma_N^2}{n_N}}} \]
Here, \( \bar{x}_F \) and \( \bar{x}_N \) are the sample means for the two groups, \( \sigma_F \) and \( \sigma_N \) are the population standard deviations, and \( n_F \) and \( n_N \) are the sample sizes for each group. By calculating the Z-score, we assess how far away, in standard deviation units, the observed difference is from the hypothesized difference (in this case 0), assuming the null hypothesis is true.

It provides a way to quantify the probability of observing such data by random chance, given that the null hypothesis of no difference is true.
P-value
The p-value is a statistical measure that helps us determine the significance of our test results. In hypothesis testing, the p-value tells us the probability of obtaining test results at least as extreme as the observed results, assuming that the null hypothesis is true. The smaller the p-value, the stronger the evidence against the null hypothesis.

In our plant study, a p-value was calculated to be approximately 0.020. This indicates there is a 2% chance of observing a Z-score of 2.057 or more extreme under the null hypothesis.

  • If the p-value is less than the chosen significance level (usually 0.05), we reject the null hypothesis.
  • If it's greater than 0.05, we do not have enough evidence to reject the null hypothesis.
The 0.020 p-value means there's significant evidence to suggest that the plant food indeed contributes to plant growth. It's an integral part of drawing meaningful conclusions from statistical testing.
Standard Normal Distribution
A standard normal distribution is a type of continuous probability distribution for a real-valued random variable. It has a mean of 0 and a standard deviation of 1. In many statistical methods, the standard normal distribution is used for calculations because it simplifies finding probabilities by using Z-scores.

The bell-shaped curve helps determine probabilities of standard normal variables, allowing us to compare different data sets on a common scale. This enables statisticians to perform hypothesis testing, like the Z-test.
  • Z-scores measure how many standard deviations an element is from the mean.
For the plant example, we calculated a Z-score of 2.057 to compare our observed mean difference against the standard normal distribution.

The corresponding probability (p-value) can be found using standard normal tables or statistical software, allowing researchers to understand how extreme their test statistics are under the null hypothesis. By visualizing this on a graph, researchers can easily see where the test statistic falls within the distribution.
Null and Alternative Hypotheses
Formulating hypotheses is a critical step in conducting a hypothesis test. It involves establishing two conflicting hypotheses that represent our tests thoroughly.

The null hypothesis \((H_0)\) is a statement that indicates no effect or no difference. In our case, it suggests that plant food does not affect the growth of the plants, which can be expressed as \( \mu_F = \mu_N \).
  • \( H_0 \) is often the hypothesis that the researcher attempts to refute.
The alternative hypothesis \((H_1)\) is contrary to the null, representing the effect the researcher wishes to test. Here, it suggests that plant food does increase plant height, expressed as \( \mu_F > \mu_N \).
  • \( H_1 \) provides a direction for the effect we're testing for – in this case, an increase in plant growth.
Testing these hypotheses involves calculating a test statistic (like the Z-score) and a corresponding p-value. Based on this, decisions can be made about rejecting or not rejecting the null hypothesis at a chosen significance level. Understanding these hypotheses is fundamental for interpreting statistical tests, as they define what you're testing and direct the decision-making process.

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Most popular questions from this chapter

Use the following information to answer the next five exercises. A doctor wants to know if a blood pressure medication is effective. Six subjects have their blood pressures recorded. After twelve weeks on the medication, the same six subjects have their blood pressure recorded again. For this test, only systolic pressure is of concern. Test at the 1% significance level. $$ \begin{array}{|l|l|l|l|l|l|}\hline \text { Patient } & {\mathbf{A}} & {\mathbf{B}} & {\mathbf{C}} & {\mathbf{D}} & {\mathbf{E}} & {\mathbf{F}} \\\ \hline \text { Before } & {161} & {162} & {165} & {162} & {166} & {171} \\\ \hline \text { After } & {158} & {159} & {166} & {160} & {167} & {169} \\\ \hline\end{array} $$ What is the sample mean difference?

Researchers interviewed street prostitutes in Canada and the United States. The mean age of the 100 Canadian prostitutes upon entering prostitution was 18 with a standard deviation of six. The mean age of the 130 United States prostitutes upon entering prostitution was 20 with a standard deviation of eight. Is the mean age of entering prostitution in Canada lower than the mean age in the United States? Test at a 1% significance level.

Use the following information to answer next five exercises. A study was conducted to test the effectiveness of a juggling class. Before the class started, six subjects juggled as many balls as they could at once. After the class, the same six subjects juggled as many balls as they could. The differences in the number of balls are calculated. The differences have a normal distribution. Test at the 1% significance level. $$ \begin{array}{|c|c|c|c|c|c|}\hline \text { Subject } & {\mathbf{A}} & {\mathbf{B}} & {\mathbf{c}} & {\mathbf{D}} & {\mathbf{E}} & {\mathbf{F}} \\\ \hline \text { Before } & {3} & {4} & {3} & {2} & {4} & {5} \\ \hline \text { After } & {4} & {5} & {6} & {4} & {5} & {7} \\ \hline\end{array} $$ What conclusion can you draw about the juggling class?

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 The league mean batting average is 0.280 with a known standard deviation of 0.06. The Rattlers and the Vikings belong to the league. The mean batting average for a sample of eight Rattlers is 0.210, and the mean batting average for a sample of eight Vikings is 0.260. There are 24 players on the Rattlers and 19 players on the Vikings. Are the batting averages of the Rattlers and Vikings statistically different?

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 windshield treatment claims to repel water more effectively. Ten windshields are tested by simulating rain without the new treatment. The same windshields are then treated, and the experiment is run again. A hypothesis test is conducted.

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