An electric device delivers a current of [tex]$15.0 A$[/tex] for 30 seconds. How many electrons flow through it?
Answer (2)
Define populations and hypotheses: The null hypothesis states equal absence frequencies across weekdays.
Calculate the expected frequency: With 100 managers, the expected frequency per day is 5 100 = 20 .
Compute the Chi-square statistic: χ 2 = 8.3 .
Compare to the critical value: Since 6.7449"> 8.3 > 6.7449 , reject the null hypothesis. The days do not occur with equal frequencies.
Explanation
Problem Setup and Objective Let's break down this problem step by step. We're given data on employee absences across a five-day work week and want to see if absences are uniformly distributed. We'll use a Chi-square goodness-of-fit test to determine this.
Hypotheses and Populations First, we need to define our populations and hypotheses.
Population 1: The set of all employee absences during a five-day workweek recorded by the managers. Population 2: The set of all employee absences during a five-day workweek expected under the assumption of uniform distribution.
Null Hypothesis ( H 0 ): The days for the highest number of absences occur with equal frequencies during a five-day work week. Alternative Hypothesis ( H r ): The days for the highest number of absences do not occur with equal frequencies during a five-day work week.
To express the alternative hypothesis with percentages, we can state that the percentages of absences for each day are not all equal to 20% (since 100%/5 = 20% ). Therefore:
H r :
Observation
Percentage
Monday
25
25%
Tuesday
24
24%
Wednesday
24
24%
Thursday
17
17%
Friday
10
10%
Expected Values and Requirements To verify the requirements for the Chi-square test, we need to check if the expected values are large enough (typically, at least 5). The expected frequency for each day, assuming a uniform distribution, is calculated as:
Number of days Total number of managers = 5 100 = 20
Since the expected value for each day is 20, which is greater than 5, the requirements for the Chi-square test are met.
Day
Expected
Monday
20
Tuesday
20
Wednesday
20
Thursday
20
Friday
20
Chi-Square Test Statistic Calculation Now, let's calculate the Chi-square test statistic. The formula is:
χ 2 = ∑ E i ( O i − E i ) 2
where O i is the observed frequency and E i is the expected frequency for each day. Plugging in the values:
χ 2 = 20 ( 25 − 20 ) 2 + 20 ( 24 − 20 ) 2 + 20 ( 24 − 20 ) 2 + 20 ( 17 − 20 ) 2 + 20 ( 10 − 20 ) 2 χ 2 = 20 25 + 20 16 + 20 16 + 20 9 + 20 100 = 20 25 + 16 + 16 + 9 + 100 = 20 166 = 8.3
So, the Chi-square test statistic is 8.3.
Degrees of Freedom and Critical Value The degrees of freedom ( df ) are calculated as the number of categories (days) minus 1:
df = 5 − 1 = 4
We are given a significance level of 15% or 0.15. We need to find the critical value of the Chi-square distribution with 4 degrees of freedom at a 15% significance level. The critical value is 6.7449.
Conclusion and Interpretation Now, we compare the calculated Chi-square test statistic (8.3) to the critical value (6.7449). Since 8.3 > 6.7449, we reject the null hypothesis.
This means there is sufficient evidence to conclude that the days for the highest number of absences do not occur with equal frequencies during a five-day work week.
Examples
Consider a retail store manager who wants to know if customer traffic is evenly distributed throughout the weekdays. By tracking the number of customers each day and performing a Chi-square test, the manager can determine if staffing levels should be adjusted to match peak traffic days, optimizing customer service and sales. This ensures resources are allocated efficiently based on actual customer behavior rather than assumptions.
By using the formula for current and charge, we calculated the total charge delivered by a device with a current of 15.0 A over 30 seconds to be 450 C. Dividing this charge by the charge of a single electron gives approximately 2.81 x 10^21 electrons that flow through the device. Thus, around 2.81 billion billion electrons pass through in this time frame.
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