/*! 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 163 It has been suggested that abnor... [FREE SOLUTION] | 91Ó°ÊÓ

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It has been suggested that abnormal male children tend to be born to older- than-average parents. Case histories of 20 abnormal males were obtained, and the ages of the 20 mothers were as follows: $$\begin{array}{lllllllllll}31 & 21 & 29 & 28 & 34 & 45 & 21 & 41 & 27 & 31 \\\43 & 21 & 39 & 38 & 32 & 28 & 37 & 28 & 16 & 39\end{array}$$ The mean age at which mothers in the general population give birth is 28.0 years. a. Calculate the sample mean and standard deviation. b. Does the sample give sufficient evidence to support the claim that abnormal male children have olderthan-average mothers? Use \(\alpha=0.05 .\) Assume ages have a normal distribution.

Short Answer

Expert verified
This solution provides the step-by-step process to calculate sample mean and sample standard deviation of the given ages. Also, provides the steps for hypothesis testing to check whether the sample provides enough evidence to support the claim that abnormal male children have older-than-average mothers.

Step by step solution

01

Calculate Sample Mean

To calculate the sample mean age of mothers, add all the ages of the mothers and divide by the number of mothers: \[ \frac{31+21+29+28+34+45+21+41+27+31+43+21+39+38+32+28+37+28+16+39}{20}\]
02

Calculate Sample Standard Deviation

To calculate the standard deviation, subtract the mean from each data point, square the result, calculate the mean of these squares, and then take the square root of that mean: \[ \sqrt{\frac{\Sigma{(x-\mu)}^2 }{N}}\] where \(\Sigma{(x-\mu)}^2\) is the sum of the squares of the differences between each data point and the mean and \(N\) is the number of observations.
03

Hypothesis Testing

State the null hypothesis \(H_0: \mu = 28\) and alternative hypothesis \(H_1: \mu > 28\). Calculate Z score \[ Z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}\] where \(\mu\) is population mean, \(\bar{x}\) is sample mean, \(s\) is standard deviation of the sample, and \(n\) is the number of samples. Use the standard normal (Z) distribution to find the critical Z value for \(\alpha=0.05\), and compare it with calculated Z score.
04

Drawing conclusion from Hypothesis Testing

If the calculated Z score is more than the critical Z value, reject the null hypothesis indicating sufficient evidence to support the claim that abnormal male children have older-than-average mothers. If not, accept the null hypothesis.

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Key Concepts

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

Sample Mean Calculation
Understanding how to calculate the sample mean is fundamental in statistics. It represents the average value in a sample data set, providing a quick glimpse into the center of the data. To find the sample mean, you sum up all the values and then divide by the number of observations. The formula is:
\[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \]
where \( \bar{x} \) is the sample mean, \( n \) is the number of observations, and \( x_i \) are the individual data points. In the context of the exercise, you added all the ages of the mothers together and then divided by the total count (20) to obtain the sample mean age.
Sample Standard Deviation
The sample standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the sample mean, while a high standard deviation indicates that the values are spread out over a wider range. To calculate the sample standard deviation (represented as \( s \)), you follow these steps:
  • Subtract the sample mean from each data point to find the deviation of each point.
  • Square each deviation.
  • Add up all the squared deviations.
  • Divide by \( n-1 \), where \( n \) is the sample size, to find the variance.
  • Take the square root of the variance to get the standard deviation.
The formula is:
\[ s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} \]
For the given exercise, you would use this process with the ages of the mothers to find out how much the ages vary from the average (sample mean) age.
Normal Distribution
The normal distribution, often called the bell curve, is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. In a normal distribution, the mean, median, and mode are all equal. This distribution is extremely important in statistics because of the central limit theorem, which states that, under many conditions, as sample size increases, the distribution of sample means approaches a normal distribution, regardless of the shape of the original data distribution.
In the case of the exercise, the ages of the mothers are assumed to follow a normal distribution, which allows us to use the Z score and the standard normal distribution table in the subsequent steps to test our hypothesis.
Z Score Calculation
The Z score is a statistical measurement that describes a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean. If a Z score is 0, it indicates that the data point's score is identical to the mean score. Calculating the Z score involves the following formula:
\[ Z = \frac{x - \mu}{\sigma} \]
In this formula, \( x \) is the value from the sample, \( \mu \) is the mean of the population, and \( \sigma \) is the population standard deviation. For sample data, where we use the sample standard deviation (\( s \)), the formula modifies to:
\[ Z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \]
where \( \bar{x} \) is the sample mean, and \( n \) is the sample size. This Z score helps us determine the likelihood of a sample mean occurring within a normal distribution.
Null and Alternative Hypothesis
Hypothesis testing in statistics is a way to test if there is enough evidence to support a certain belief (alternative hypothesis) about a parameter compared to what has been generally accepted (null hypothesis). The null hypothesis (\( H_0 \)) is a statement of no effect or no difference and is usually the hypothesis that experimenters try to disprove. On the other hand, the alternative hypothesis (\( H_1 \) or (\( H_a \))) expresses a specific claim that there is a significant difference, relationship, or effect.
The process of hypothesis testing involves stating both hypotheses, determining a level of significance (\( \alpha \)), and finding the critical value (Z score or t score) to either reject or fail to reject the null hypothesis. In your exercise, the null hypothesis states that the mean age of mothers of abnormal male children is equal to 28 (\( H_0: \mu = 28 \)), and the alternative hypothesis suggests that the mean age is greater than 28 (\( H_1: \mu > 28 \)).

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Most popular questions from this chapter

You are testing the hypothesis \(p=0.7\) and have decided to reject this hypothesis if after 15 trials you observe 14 or more successes. a. If the null hypothesis is true and you observe 13 successes, which of the following will you do? (1) Correctly fail to reject \(H_{o} .\) (2) Correctly reject \(H_{o} .(3)\) Commit a type I error. (4) Commit a type II error. b. Find the significance level of your test. c. If the true probability of success is \(1 / 2\) and you observe 13 successes, which of the following will you do? (1) Correctly fail to reject \(H_{o} .\) (2) Correctly reject \(H_{o} .(3)\) Commit a type I error. (4) Commit a type II error. d. Calculate the \(p\) -value for your hypothesis test after 13 successes are observed.

The uniform length of nails is very important to a carpenter-the length of the nails being used are matched to the materials being fastened together, thereby making a small standard deviation an important property of the nails. A sample of 35 randomly selected 2-inch nails is taken from a large quantity of Nails, Inc.'s, recent production run. The resulting length measurements have a mean length of 2.025 inches and a standard deviation of 0.048 inch. a. Determine whether an assumption of normality is reasonable. Explain. b. Is the sample evidence sufficient to reject the idea that the nails have a mean length of 2 inches? Use \(\alpha=0.05\). c. Is there sufficient evidence, at the 0.05 level, to show that the length of nails from this production run has a standard deviation greater than the advertised 0.040 inch? d. Write a short report outlining the findings and recommendations as to whether or not the carpenter should use these nails for an application that requires 2 -inch nails.

Julia Jackson operates a franchised restaurant that specializes in soft ice cream cones and sundaes. Recently she received a letter from corporate headquarters warning her that her shop is in danger of losing its franchise because the average sales per customer have dropped "substantially below the average for the rest of the corporation." The statement may be true, but Julia is convinced that such a statement is completely invalid to justify threatening a closing. The variation in sales at her restaurant is bound to be larger than most, primarily because she serves more children, elderly, and single adults rather than large families who run up big bills at the other restaurants. Therefore, her average ticket is likely to be smaller and exhibit greater variability. To prove her point, Julia obtained the sales records from the whole company and found that the standard deviation was \(2.45\)dollar per sales ticket. She then conducted a study of the last 71 sales tickets at her store and found a standard deviation of \(2.95\)dollar per ticket. Is the variability in sales at Julia's franchise, at the 0.05 level of significance, greater than the variability for the company?

It is claimed that the students at a certain university will score an average of 35 on a given test. Is the claim reasonable if a random sample of test scores from this university yields \(33,42,38,37,30,42 ?\) Complete a hypothesis test using \(\alpha=0.05 .\) Assume test results are normally distributed. a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

You are interested in comparing the null hypothesis \(p=0.8\) against the alternative hypothesis \(p<0.8 .\) In 100 trials you observe 73 successes. Calculate the \(p\) -value associated with this result.

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