Question

A sample of salary offers (in thousands of dollars) given to management majors is: 48, 51, 46, 52, 47, 48, 47, 50, 51, and 59. Using this data to obtain a 95 percent confidence interval resulted in an interval from 47.19 to 52.61. True or False: The confidence interval obtained is valid only if the distribution of the population of salary offers is normal.

Answer 1

Answer:

Step-by-step explanation:

number of samples, n = 10

Mean = (48 + 51 + 46 + 52 + 47 + 48 + 47 + 50 + 51 + 59)/10 = 49.9

Standard deviation = √(summation(x - mean)/n

Summation(x - mean) = (48 - 49.9)^2 + (51 - 49.9)^2 + (46 - 49.9)^2+ (52 - 49.9)^2 + (47 - 49.9)^2 + (48 - 49.9)^2 + (47 - 49.9)^2 + (50 - 49.9)^2 + (51 - 49.9)^2 + (59- 49.9)^2 = 128.9

Standard deviation = √128.9/10 = 3.59

Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Margin of error = z × s/√n

Where

s = sample standard deviation

From the information given, the population standard deviation is unknown and the sample size is small, hence, we would use the t distribution to find the z score

In order to use the t distribution, we would determine the degree of freedom, df for the sample.

df = n - 1 = 10 - 1 = 9

Since confidence level = 95% = 0.95, α = 1 - CL = 1 – 0.95 = 0.05

α/2 = 0.05/2 = 0.025

the area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 1 - 0.025 = 0.975

Looking at the t distribution table,

z = 2.262

Margin of error = 2.262 × 3.59/√10

= 2.57

the lower limit of this confidence interval is

49.9 - 2.57 = 47.33

the lower limit of this confidence interval is

49.9 + 2.57 = 52.47

So it is false