(10 points) starting salaries of 64 college graduates who have taken a statistics course have a mean of $42,500 with a standard deviation of $6,800. find an 90% confidence interval for (note: do not use commas or dollar signs in your answers. round each bound to three decimal places.) lower-bound: 41101.87 upper-bound: 43898.13



41101.750 to 43898.250

Step-by-step explanation:

Using this formula X ± Z (s/√n)

Where

X = 42500 Mean

S = 6800 Standard Deviation

n = 64 Number of observation

Z = 1.645 The chosen Z-value from the confidence table below

Confidence IntervalZ

80%. 1.282

85% 1.440

90%. 1.645

95%. 1.960

99%. 2.576

99.5%. 2.807

99.9%. 3.291

Substituting these values in the formula

Confidence Interval (CI) = 42500 ± 1.645(6800/√64)

CI = 42500 ± 1.645(6800/8)

CI = 42500 ± 1.645(850)

CI = 42500 ± 1398.25

CI = 42500+1398.25 ~. 42500-1398.25

CI = 43898.25 ~ 41101.75

In other words the confidence interval is from 41101.750 to 43898.250

3 is greater then both of those numbers

ans789

step-by-step explanation: