Complete question :
According to the National Association of Realtors, it took an average of three weeks to sell a home in 2017. Suppose data for the sale of 39 randomly selected homes sold in Greene County, Ohio, in 2017 showed a sample mean of 3.6 weeks with a sample standard deviation of 2 weeks. Conduct a hypothesis test to determine whether the number of weeks until a house sold in Greene County differed from the national average in 2017. Useα = 0.05for the level of significance, and state your conclusion
Answer:
H0 : μ = 3
H1 : μ ≠ 3
Test statistic = 1.897
Pvalue = 0.0653
fail to reject the Null ; Hence, we conclude that their is no significant to accept the claim that number I weeks taken to sell a house differs.
Step-by-step explanation:
Given :
Sample size, n = 40
Sample mean, x = 3.6
Population mean, μ = 3
Standard deviation, s = 2
The hypothesis :
H0 : μ = 3
H1 : μ ≠ 3
The test statistic :
(xbar - μ) ÷ (s/√n)
(3.6 - 3) / (2/√40)
0.6 / 0.3162277
Test statistic = 1.897
Using T test, we can obtain the Pvalue from the Test statistic value obtained :
df = n - 1; 40 - 1 = 39
Pvalue(1.897, 39) = 0.0653
Decison region :
If Pvalue ≤ α ; Reject the null, if otherwise fail to reject the Null.
α = 0.05
Pvalue > α ; We fail to reject the Null ; Hence, we conclude that their is no significant to accept the claim that number I weeks taken to sell a house differs.
Well the are six sides and numbers on the dices and two sides on the coin so there would be 8 outcomes
0.17.
Look at the numbers they have the same. They both start with 0.01...but the next number in the first one is 7, which is greater than 6.
we have
step 1
Multiply by 2 both sides
step 2
Group terms
Combine like terms
divide by 2 both sides
Verify
substitute the value of x in the original expression
is ok
the value of x satisfy the equation
