1.A random sample of 70 observations produced a mean of x=31.5x from a populatio
ID: 3134804 • Letter: 1
Question
1.A random sample of 70 observations produced a mean of x=31.5x from a population with a normal distribution and a standard deviation =2.45
(a) Find a 95% confidence interval for
(b) Find a 99% confidence interval for
c) Find a 90% confidence interval for
2. Starting salaries of 130 college graduates who have taken a statistics course have a mean of $44,609. Suppose the distribution of this population is approximately normal and has a standard deviation of $10,228.
Using an 85% confidence level, find both of the following:
(a) The margin of error:
(b) The confidence interval for the mean :
3. A random sample of nn measurements was selected from a population with standard deviation =15 and unknown mean . Calculate a 90% confidence interval for for each of the following situations:
(a) n=55, x=84.8
(b) n=85, x=84.8
(c) n=115, x=84.8
(d) In general, we can say that for the same confidence level, increasing the sample size ____________ the margin of error (width) of the confidence interval
4. What is the value of t*, the critical value of the t distribution for a sample of size 18, such that the probability of being greater than t* is 1%?
5.What is the value of t*, the critical value of the t distribution for a sample of size 14, such that the probability of being greater than t* or less than -t* is 1%?
Please and thank you!!!
Explanation / Answer
Answer to question# 1)
Formula of confidence interval is :
x bar - z* SE , xbar + z*SE
.
We got x bar = 31.5
SE = 2.45 / sqrt(70)
SE = 0.2928
.
Answer to part a)
For 95% confidence interval , z = 1.96
.
On plugging the values weget:
31.5 - 1.96 * 0.2928 , 31.5 + 1.96 *0.28928
30.9261 , 32.067
.
Answer to part b)
Z value for 99% confidence level is: 2.58
31.5 - 2.58 *0.2928 , 31.5 + 2.58 *0.2928
30.7446 , 32.2554
.
Answer to part c)
Z value for 90% confidence interval is 1.645
31.5 -1.645 *0.2928 , 31.5 +1.645 *0.2928
31.0183 , 31.9817
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