Solution :
Given that,
Point estimate = sample mean = = 18400
sample standard deviation = s = 10600
sample size = n = 40
Degrees of freedom = df = n - 1 = 40 - 1 = 39
At 80% confidence level the t is ,
= 1 - 80% = 1 - 0.80 = 0.20
/ 2 = 0.20 / 2 = 0.10
t /2,df = t0.10 , 39 = 1.304
Margin of error = E = t/2,df * (s /n)
= 1.304 * (10600 / 40)
= 2186
The 80% confidence interval estimate of the population mean is,
- E < < + E
18400 - 2186 < < 18400 + 2186
16214 < < 20586
please solve College tuition: A simple random sample of 40 colleges and universities in the United...
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