College tuition: a simple random sample of 35 colleges and universities in the united states has a mean tuition of $19,400 with a standard deviation of $10,700. Construct a 95% confidence interval for the mean tuition for all colleges and universities in the United States. round the answers to the nearest whole number.
A 95% confidence interval for the mean tuition for all colleges and universities is __________________< mean < _____________________
Solution :
Given that,
= $19400
s = $10700
n = 35
Degrees of freedom = df = n - 1 = 35 - 1 = 34
At 95% confidence level the t is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
t /2,df = t0.025,34 = 2.032
Margin of error = E = t/2 * (s /n) = 2.032 * (10700/ 35) = 3675
A 95% confidence interval for the mean tuition for all colleges and universities is
- E < < + E
19400 - 3675 < < 19400 + 3675
$15725 < < $23075
College tuition: a simple random sample of 35 colleges and universities in the united states has...
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please solve College tuition: A simple random sample of 40 colleges and universities in the United States has a mean tuition of $18,400 with a standard deviation of $10,600. Construct an 80% confidence interval for the mean tuition for all colleges and universities in the United States. Round the answers to the nearest whole number. An 80% confidence interval for the mean tuition for all colleges and universities is ㄨ 幻
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