The students at Littlewood Regional High School cut an average of 3.3 classes per week. A random sample of 117 seniors averages 3.8 cuts per week, with a standard deviation of 0.53. Are seniors...

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This question can be solved in 2 ways, leading to different answers. First is
using z (normal dist) since n = 117 ≥ 30 (large), and population std dev σ can
be approximated by sample std dev (sx). Second method is to use t distribution
since population std dev is unknown.
I am covering both, but you need to select first if you were taught to use z for
large samples, otherwise choose second.
Test: µ 6= 3.3 with normal distn
Hypothesis test:
H0 : µ = 3.3 (Null Hypothesis)
Ha : µ 6= 3.3 (Alternative Hypothesis, also called H1)
This is two tailed test.
Since sample size is large, sample standard deviation can be taken as an approx-
imation of population standard deviation.
x = 3.8
σ = 0.53
n = 117
Significance Level, α = 0.05 (If no value is given, we take level of 0.05)
Test statistic z? = x− µ
σ/
√
n
Test Statistic, z? = 3.8− 3.3
0.53/
√
117
≈ 10.20439 ≈ 10.20
Test Statistic, z? = 10.20
P-value Approach:
P-value is in direction of Alternative hypothesis.
Since Ha : µ 6= 3.3, P-value= P (Z < −10.2 or Z > 10.2)
P-value is Area on left of −10.2 and on right of test statistic = 10.2
0.0000
−10.2
0
0.0000
test statistic = 10.2
P-value = 2 ∗ P (Z < −10.2) = 2 ∗ 0 =0.0000 (from z-table)
P-value = 0.0000
Rejection Rule: Reject H0 if p-value < α
Using excel function 2*normdist( -abs( (3.8 - 3.3)/( 0.53/sqrt(117))),0,1, TRUE) or TI’s func-
tion 2*normalcdf( 1E-99, -abs( (3.8 - 3.3)/( 0.53/sqrt(117))) ) answer is: 0
Decision: Since 0.0000 < 0.05, we reject the null...