I begin with a three-digit positive integer. I divide it by 9 and then subtract 9 from the answer. My final answer is also a thr
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Answer:
a) sample mean = 63
b) 95% confidence interval for the population mean wage (including benefits) for these employees is ( 60.58, 65.42 )
c) sample size is 26
Step-by-step explanation:
given the data in the question
a) estimate of the population mean
sample mean = ($63.00 per hour) = 63
b)
given that; standard deviation σ = 6.11, sample size n = 27
df = n-1 = 27 - 1 = 26
now at 95% CI, t will be;
∝ = 1 - 95% = 0.05, = 0.05/2 = 0.025
=
= 2.056
Margin of Error E = × (σ/√n)
Margin of Error E = 2.056 × (6.11/√27)
Margin of Error E = 2.056 × 1.17587
Margin of Error E = 2.4175 ≈ 2.42
CI estimate of the population mean will be; ( 95% )
- E < μ <
+ E
we substitute
63 - 2.42 < μ < 63 + 2.42
60.58 < μ < 65.42
Therefore, 95% confidence interval for the population mean wage (including benefits) for these employees is ( 60.58, 65.42 )
c)
At 90% confidence level and Margin of Error of 2
sample size n = ?
∝ = 1-90% = 0.10, ∝/2 = 0.10/2 = 0.05
=
= 1.645
Sample size n = ( × σ/ / E )²
we substitute
Sample size n = (1.645×6.11 / 2)²
n = (10.05095 / 2)²
n = ( 5.025475)²
n = 25.255
number of employed judges cant be decimal,
Therefore, sample size is 26