OPTION A
The average temperature in Fairbanks Alaska in February is 4F below zero write this temperature as an integer
2 answers:
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OPTION A
Answer:
Step-by-step explanation:
This is your equation, right? Just checking
We would need the sample standard deviation for our hypothesis test.
STEP 1: Work out the mean of data set
μ = Sum of values in the set ÷ Number of values in the set
μ = 1686 ÷ 14
μ = 120.4 (rounded to one decimal place)
STEP 2: Subtract the mean from each value in the data set, and then square each answer. The table attached below shows the details of the calculation
STEP 3: Add the answers in STEP 2, then divide by 13.
Note: we divide by 13 instead of 14 because our data set is a sample set.
(∑x₁-μ)/13 = 1253.44÷13 = 96.42
The value obtained in STEP 3 is the variance. To obtain the standard deviation, we square root the variance
s = √96.42 = 9.819 (rounded to three decimal places)
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We want to test the claim that the standard deviation of IQ scores of college students is less than 15. Putting this in the form of hypotheses we get:
H₀: σ=15
H₁: σ<15
The significance level, α, is 0.10
The degrees of freedom, v, is 14 - 1 = 13
Reading from the Chi² table for α = 0.1 and degree of freedom = 13, we have the critical value ≥19.812
The value of the test statistic is given as [(n-1)²s²] ÷ σ²
We have n = 14, s = 9.819, and σ = 15
test statistic = [(13)²(9.819)²] ÷ 15² = 72.417
72.417 is in the critical region (which is values ≥ 19.812) so the result is significant and H₀ is rejected. The standard deviation of IQ scores of college students is less than 15.
STEP 1: Work out the mean of data set
μ = Sum of values in the set ÷ Number of values in the set
μ = 1686 ÷ 14
μ = 120.4 (rounded to one decimal place)
STEP 2: Subtract the mean from each value in the data set, and then square each answer. The table attached below shows the details of the calculation
STEP 3: Add the answers in STEP 2, then divide by 13.
Note: we divide by 13 instead of 14 because our data set is a sample set.
(∑x₁-μ)/13 = 1253.44÷13 = 96.42
The value obtained in STEP 3 is the variance. To obtain the standard deviation, we square root the variance
s = √96.42 = 9.819 (rounded to three decimal places)
----------------------------------------------------------------------------------------------------------------
We want to test the claim that the standard deviation of IQ scores of college students is less than 15. Putting this in the form of hypotheses we get:
H₀: σ=15
H₁: σ<15
The significance level, α, is 0.10
The degrees of freedom, v, is 14 - 1 = 13
Reading from the Chi² table for α = 0.1 and degree of freedom = 13, we have the critical value ≥19.812
The value of the test statistic is given as [(n-1)²s²] ÷ σ²
We have n = 14, s = 9.819, and σ = 15
test statistic = [(13)²(9.819)²] ÷ 15² = 72.417
72.417 is in the critical region (which is values ≥ 19.812) so the result is significant and H₀ is rejected. The standard deviation of IQ scores of college students is less than 15.