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Answer:
The 99% confidence interval of the population standard deviation is 1.7047 < σ < 7.485
Step-by-step explanation:
Confidence interval of standard deviation is given as follows;
s =
Where:
= Sample mean
s = Sample standard deviation
n = Sample size = 7
χ = Chi squared value at the given confidence level
= ∑x/n = (62 + 58 + 58 + 56 + 60 +53 + 58)/7 = 57.857
The sample standard deviation s = = 2.854
The test statistic, derived through computation, = ±3.707
Which gives;
1.7047 < σ < 7.485
The 99% confidence interval of the population standard deviation = 1.7047 < σ < 7.485.
This question is incomplete, the complete question;
The American Psychological Associated conducted a survey in 2003 of psychologists to estimate the average incomes for psychologists based on their academic degree and years of experience. 36 psychologists responded who had a just received their masters degree in 2003. The average was $48,800 with a standard deviation of $16,800.
The 95% Confidence Interval is ($, $ )
Answer:
at 95%, confidence interval is ( $43312, $54288 )
Step-by-step explanation:
Given the data in the question;
sample size n = 36
mean x' = 48,800
standard deviation σ = 16,800
at 95% confidence level;
∝ = 1 - 95% = 1 - 0.95 = 0.05
∝/2 = 0.05 / 2 = 0.025
z critical value, = 1.96
now, confidence interval for population mean is given by the formula
x' ± ( σ/√n )
so we substitute
⇒ 48800 ± 1.96( 16800/√36 )
⇒ 48800 ± 1.96( 2800 )
⇒ 48800 ± 5488
⇒ 48800 - 5488, 48800 + 5488
⇒ ( 43312, 54288 )
Therefore, at 95%, confidence interval is ( $43312, $54288 )