A real estate developer is considering investing in a shopping mall on the outskirts of Atlanta, Georgia. Three parcels of land are being evaluated. Of particular importance is the income in the area surrounding the proposed mall. A random sample of four families is selected near each proposed mall. Following are the sample results. At the 0.05 significance level, can the developer conclude there is a difference in the mean income
Answers
Answer:
Step 1: The null and alternative hypothesis for the one way analysis of variance is,
H₀ : There is no difference in the means of each area.
H₁ : At least one mean is different.
The objective of this question is to test whether the income of the three areas are significantly different.
Step 2: The problem states to use 0.05 significance level.
Step 3: The test statistic follows the F-distribution.
Step 4: Decision rule is based on the critical value.
Numerator degrees of freedom is,
K - 1 = 3 - 1
= 2
Denominator degrees of freedom is 2.
n - k = 12 - 3
= 9
Using the F-distribution table for ∝=0.05 with numerator and denominator degrees freedom, the critical value is 4.26. The decision rule is to reject H₀ if the computed value of F exceeds 4.26.
Step 5: Calculations.
Using Excel follow the steps mentioned below.
1. Import or type the data into spreadsheet.
2. Data--------> Data Analysis----------> Anova: Single Factor, click OK.
3. Select Input Range, make a tick mark on Labels in the first row, enter 0.05 for alfa.
4. Click OK.
The obtained output for the one-way ANOVA is,
[ find the figure in attachment]
From the obtained output, the computed test statistic value, F = 14.18 , which is greater than the critical value of 4.26. So, reject H₀
Step 6: Interpretation. The average income of each area is different. It can be concluded that the average income is significantly different .
Final answer:
To determine if there is a difference in the mean income between three proposed mall locations, perform an analysis of variance (ANOVA) test.
Explanation:
To determine if there is a difference in the mean income between the three proposed mall locations, we can perform an analysis of variance (ANOVA) test. This test assesses whether there is a statistically significant difference among the means of multiple groups.
First, calculate the sample mean income for each location. Then, compute the sum of squares within groups (SSW) and the sum of squares between groups (SSB). Finally, use these values to calculate the F-statistic and compare it to the critical value from the F-distribution table at the 0.05 significance level to make a conclusion.
If the computed F-statistic is greater than the critical value, we reject the null hypothesis and conclude that there is a difference in the mean income.
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Related Questions
Mario has drawn a plan of his bedroom on 1 cm square paper. His en-suite shower cubicle measure. 1m x1m, give the scale of his drawing as ratio ___ : ___.What are the actual dimension of his bed ____ m x __ m
Based on the data, select the most reasonable prediction about the height ofsecond-grade students.A. The typical second-grade student is less than 45 inches tall.B. The typical second-grade student is about 45 inches tall.C. The typical second-grade student is more than 50 inches tall.D. The typical second-grade student is about 48 inches tall.
Please help me with this homework equation I don’t understand and need help someone please do step by step
Pls help!!!!! i’m on a timer i’ll mark you brainliest!!!
Answers
Answer:
36
Step-by-step explanation:
well you could add 24+12 which equals 36
Answers
Answer:
10.48
Step-by-step explanation:
Answer:
10.48 mi
Step-by-step explanation:
2.62 times 4= 10.48
x=
Answers
Answer:
x = 80
Step-by-step explanation:
9+1=10 so x/8 must equal 10.
80/8 = 10 and subract by 9 = 1
2. -1• X/-1 = 1• -1
( multiple -1 by both sides)
3. X= -1
I hope this helps
Answers
Answer:
N → S (If national elections deteriorate into TV popularity contests, then smooth-talking morons will get elected.)
¬N → ¬S ( Therefore, if national elections do not deteriorate into TV popularity contests, then smooth-talking morons will not get elected.)
Step-by-step explanation:
¬ is a symbol of negation
→ is a symbol that represent "if... then..." scenario
Answers
Answer:
There is a 32.22% probability that at least 3 flights arrive late.
Step-by-step explanation:
For each flight, there are only two possible outcomes. Either it arrives on time, or it arrives late. This means that we can solve this problem using binomial probability concepts.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinatios of x objects from a set of n elements, given by the following formula.
And is the probability of X happening.
In this problem, we have that:
There are 10 flights, so .
A success in this case is a flight being late. 80% of its flights arriving on time, so 100%-80% = 20% arrive late. This means that .
(a) Find the probability that at least 3 flights arrive late.
Either less than 3 flights arrive late, or at least 3 arrive late. The sum of these probabilities is decimal 1. This means that:
In which
So
Finally
There is a 32.22% probability that at least 3 flights arrive late.
Final answer:
The problem is solved by calculating the probability of the complementary event (0,1,2 flights arriving late) using the binomial distribution, then subtracting this from 1 to find the probability of at least 3 flights arriving late.
Explanation:
This problem is typically solved by using a binomial probability formula, which is used when there are exactly two mutually exclusive outcomes of a trial, often referred to as 'success' and 'failure'.
Here, our 'success' is a flight arriving late. The probability of success, denoted as p, is thus 20% or 0.2 (since 80% arrive on time, then 100%-80% = 20% arrive late). The number of trials, denoted as n, is 10 (the number of randomly selected flights).
We want to find the probability that at least 3 flights arrive late, in other words, 3,4,...,10 flights arrive late. The problem can be solved easier by considering the complementary event: 0,1,2 flights arrive late. Then subtract the sum of these probabilities from 1.
The binomial probability of exactly k successes in n trials is given by:
P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))
Where C(n, k) is the binomial coefficient, meaning choosing k successes from n trials.
We calculate like so:
P(X=0) = C(10, 0) * (0.2)^0 * (0.8)^10
P(X=1) = C(10, 1) * (0.2)^1 * (0.8)^9
P(X=2) = C(10, 2) * (0.2)^2 * (0.8)^8
Sum these up and subtract from 1 to get the probability that at least 3 flights arrive late. This gives the solution to the question.
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the measure of the second smallest angle.
118°
131°
132.5°
142°
None of these answers are correct.
Answers
Answer: 132.5 degrees (choice C)
=====================================================
Explanation:
The interior angles are consecutive numbers such as 1,2,3,... or 7,8,9... and so on. The gap between any two adjacent neighbors is 1.
For any polygon with n sides, the interior angles add up to 180(n-2)
We have n = 8 sides so the interior angles sum to 180(n-2) = 180(8-2) = 1080 degrees.
Any octagon has its interior angles add up to 1080 degrees.
-----------------------------
Let x be the smallest angle. The next angle up is x+1. After that is x+2 and so on until we reach x+7 as the 8th angle.
Add up those 8 angles, set the sum equal to 1080 and solve for x
x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)+(x+7) = 1080
8x+28 = 1080
8x = 1080-28
8x = 1052
x = 1052/8
x = 131.5
This is the smallest angle. The next angle up or the second smallest angle is x+1 = 131.5+1 = 132.5 degrees (choice C)