91Ó°ÊÓ

Suppose "sleep-trained" babies (allowed to cry themselves to sleep) wake up an average of 1.2 times a night with a standard deviation of o. 3 times, while untrained babies wake up an average of 1.8 times a night with a standard deviation of 0.5 times. In a random sample of 80 babies, half of which are sleep-trained, what is the probability that the untrained babies in the sample wake up an average number of times greater than 0.75 more than the average of the sleep-trained babies? (A) \(P\left(z>\frac{0.75-0.6}{\sqrt{\frac{(0.5)^{2}}{40}-\frac{(0.3)^{2}}{40}}}\right)\) (B) \(P\left(z>\frac{0.75-0.6}{\sqrt{\frac{(0.5)^{2}}{40}-\frac{(0.3)^{2}}{40}}}\right)\) (C) \(P\left(z>\frac{0.75-0.6}{\sqrt{\frac{0.5}{40}-\frac{0.3}{40}}}\right)\) (D) \(P\left(z>\frac{0.75-0.6}{\sqrt{\frac{0.5}{40}-\frac{0.3}{40}}}\right)\) (E) \(P\left(z>\frac{0.75-0.6}{\frac{0.5}{\sqrt{40}}+\frac{0.3}{\sqrt{40}}}\right)\)

Step by step solution

01

Define the Variables

Let the average wake-up times for sleep-trained and untrained babies be denoted as \(\bar{X}_1 = 1.2\) and \(\bar{X}_2 = 1.8\), respectively. Standard deviations are \(\text{σ}_1 = 0.3\) for sleep-trained and \(\text{σ}_2 = 0.5\) for untrained babies.

02

Sample Sizes

The sample sizes for both groups are equal, which means \(n_1 = n_2 = 40\).

03

Difference in Sample Means

Determine the difference we're interested in: \(\bar{X}_2 - \bar{X}_1 > 0.75\). This simplifies to finding the probability that \(\bar{X}_2 - \bar{X}_1 = 0.75 + 0.6 = 1.35\).

04

Calculate Standard Error

Calculate the standard error for the difference in sample means: \(\text{SE} = \sqrt{\frac{(\text{σ}_1)^2}{n_1} + \frac{(\text{σ}_2)^2}{n_2}} = \sqrt{\frac{(0.3)^2}{40} + \frac{(0.5)^2}{40}}\).

05

Simplify the Standard Error

Simplify to find \(\text{SE}: \sqrt{\frac{0.09}{40} + \frac{0.25}{40}} = \sqrt{\frac{0.34}{40}} = \sqrt{0.0085} = 0.092\).

06

Form the Z-Score

Form the Z-score: \(z = \frac{0.75}{0.092}\). From calculation, \(z = 8.15\).

07

Determine the Corresponding Probability

Convert the Z-score into a probability using a Z-table. For \(z > 8.15\), the probability is extremely small and approximates to 0. Essentially, highly unlikely.

08

Identify the Option

Upon inspecting the given options, it is clear that option (A) matches our calculated standard error and Z-score formation.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Sample Mean

The sample mean is the average of a set of observations. It's calculated by summing all the values and dividing by the number of observations. For instance, if you have 10 sample data points, say 1, 2, 3, ..., 10, the sample mean \( \bar{X} \) would be calculated as \[ \bar{X} = \frac{1+2+3+...+10}{10}. \] In our problem, sleep-trained babies have a sample mean of 1.2 wake-ups per night, and untrained babies have a sample mean of 1.8 wake-ups per night.

Standard Deviation

Standard deviation measures the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values are close to the mean, while a high standard deviation indicates that the values are spread out. It's calculated using the formula: \[ \sigma = \sqrt{\frac{\sum{(X - \bar{X})^2}}{N}} \, \] where \( \sigma \) is the standard deviation, \( X \) is each value, \( \bar{X} \) is the mean of the values, and \( N \) is the number of values. In our example, the standard deviation is 0.3 for sleep-trained babies and 0.5 for untrained babies. These values tell us that untrained babies have more variability in wake-ups per night.

Z-score Calculation

The Z-score tells us how many standard deviations an element is from the mean. It's calculated using the formula: \[ z = \frac{X - \mu}{\sigma}, \] \where \( X \) is the value, \( \mu \) is the population mean, and \( \sigma \) is the standard deviation.
In our problem, we are calculating a Z-score to determine the probability that untrained babies wake up 0.75 more times than sleep-trained babies. The Z-score calculation in the exercise is \[ z = \frac{0.75 - 0.6}{\sqrt{\frac{(0.5)^2}{40} + \frac{(0.3)^2}{40}}} \] which simplifies to \[ z = \frac{0.75}{0.092} = 8.15. \] This Z-score (8.15) is extremely high, indicating that such an event is highly unlikely.

Probability

Probability is a measure of how likely an event is to occur. It ranges from 0 (impossible) to 1 (certain). In our problem, we use the Z-score to find the probability. Since we calculated the Z-score as 8.15, we look it up in a Z-table.
Z-tables give the probability that a statistic is less than a given value. For a Z-score of 8.15, the probability is nearly 1, meaning it's almost certain that a statistic is less than 8.15. Consequently, the complementary probability (the event occurring) is nearly 0, indicating that it's highly unlikely for the untrained babies to wake up more than 0.75 times more than sleep-trained ones.