For randomly selected adults, iq scores are normally distributed with a standard deviation of 15. the scores of 14 randomly selected college students are listed below. use a 0.10 significance level to test the claim that the standard deviation of iq scores of college students is less than 15. round the sample standard deviation to three decimal places. 115 128 107 109 116 124 135 127 115 104 118 126 129 133

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.

Value is 6 i think k