Question
In: Statistics and Probability
Forty items are randomly selected from a population of 450 items. The sample mean is 29,...
Forty items are randomly selected from a population of 450 items. The sample mean is 29, and the sample standard deviation 2. Develop a 95% confidence interval for the population mean. (Round the final answers to 2 decimal places.)
The confidence interval is between and .
Solutions
Expert Solution
Solution :
Given that,
= 29
s =2
n = 450
Degrees of freedom = df = n - 1 =450 - 1 = 449
a ) At 95% confidence level the t is ,
= 1 - 95% = 1 - 0.95 = 0.05
/
2= 0.05 / 2 = 0.025
t
/2,df = t0.025,449 = 1.965 ( using student t
table)
Margin of error = E = t/2,df
* (s /
=1.965 * ( 2/
= 0.19
The 95% confidence interval estimate of the population mean is,
- E < 
<
+ E
29 -0.19 <
<29 + 0.19
28.81 <
< 29.19
(28.81 ,29.19 )
orchestra answered 3 years agoRelated Solutions
Sixty items are randomly selected from a population of 660 items. The sample mean is 38,...
Sixty items are randomly selected from a population of 660
items. The sample mean is 38, and the sample standard deviation 3.
Develop a 98% confidence interval for the population mean.
(Round the final answers to 2 decimal
places.)
The confidence interval is between
and
.
Thirty-three items are randomly selected from a population of 350 items. The sample mean is 32,...
Thirty-three items are randomly selected from a population of
350 items. The sample mean is 32, and the sample standard deviation
6. Develop a 90% confidence interval for the population mean.
(Round the t-value to 3 decimal places. Round the
final answers to 2 decimal places.)
The confidence interval is between and
Thirty-four items are randomly selected from a population of 260 items. The sample mean is 33,...
Thirty-four items are randomly selected from a population of 260
items. The sample mean is 33, and the sample standard deviation 6.
Develop a 95% confidence interval for the population mean.
(Round the t-value to 3 decimal places. Round the
final answers to 2 decimal places.)
The confidence interval is between and.
A sample of 37 observations is selected from a normal population. The sample mean is 29,...
A sample of 37 observations is selected from a normal
population. The sample mean is 29, and the population standard
deviation is 5. Conduct the following test of hypothesis using the
0.05 significance level. H0: μ ≤ 26 H1: μ > 26 Is this a one- or
two-tailed test? "One-tailed"—the alternate hypothesis is greater
than direction. "Two-tailed"—the alternate hypothesis is different
from direction. What is the decision rule? (Round your answer to 2
decimal places.)
A sample of n=28 individuals is randomly selected from a population with a mean of µ=...
A sample of n=28 individuals is randomly selected from a
population with a mean of µ= 63, and a treatment is administered to
the individuals in the sample. After treatment, the sample mean is
found to be M=68.
a. If the sample variance = 96, are the data sufficient to reject
the null and conclude that the treatment has a significant effect
using a two-tailed test with alpha = .05? Be sure to show all
formulas with symbols (and plug...
A manager records the repair cost for 29 randomly selected washers. A sample mean of $84.87...
A manager records the repair cost for 29 randomly selected
washers. A sample mean of $84.87 and sample standard deviation of
$12.34 are subsequently computed. Assume the population is normally
distributed. Determine the upper endpoint of a 95% confidence
interval estimate of the true mean repair cost.
A)84.87
B)88.83
C)86.83
D)30.89
E)89.56
A manager records the repair cost for 29 randomly selected washers. A sample mean of $84.87...
A manager records the repair cost for 29 randomly selected
washers. A sample mean of $84.87 and sample standard deviation of
$12.34 are subsequently computed. Assume the population is normally
distributed. Determine the upper endpoint of a 95% confidence
interval estimate of the true mean repair cost.
A. 86.83
B. 84.87
C. 89.56
D. 30.89
E. 88.83
A sample of 10 measurements, randomly selected from a normally distributed population, resulted in a sample...
A sample of 10 measurements, randomly selected from a
normally distributed population, resulted in a sample mean=5.2, and
sample standard deviation=1.8. Using ,alpha=0.01 test the null
hypothesis that the mean of the population is 3.3 against the
alternative hypothesis that the mean of the population < 3.3, by
giving the following:
degrees of freedom =9
critical t value
the test statistic
A sample of 15 measurements, randomly selected from a normally distributed population, resulted in a sample...
A sample of 15 measurements, randomly selected from a normally
distributed population, resulted in a sample mean,
x¯¯¯=6.1 and sample standard deviation s=1.92.
Using α=0.1, test the null hypothesis that μ≥6.4
against the alternative hypothesis that μ<6.4 by giving
the following.
a) The number of degrees of freedom is: df=
.
b) The critical value is: tα= .
c) The test statistic is: ttest=
From a normal population, a sample of 39 items is taken. The sample mean is 12...
From a normal population, a sample of 39 items is taken. The
sample mean is 12 and the sample standard deviation is 2. Construct
a 99% confidence interval for the population mean.
ADVERTISEMENT
ADVERTISEMENT
Latest Questions
- During the first month of operations ended May 31, Big Sky Creations Company produced 55,500 designer...
- Write a program that input two integer number and display results by performing the (+, -,...
- 46. At the beginning of the month, the Painting Department of Skye Manufacturing had 20,000 units...
- Consider how each of the situations described below would affect the experimental value determined for the...
- A 110.0 −mL sample of a solution that is 3.0×10−3 M in AgNO3 is mixed with...
- A water stream at a mass flow rate of 0.3 kg/s and temperature of 70°C flows...
- PLEASE ANSWER EACH QUESTION 1 thru 4 Question 1. In a market with perfectly competitive firms,...
ADVERTISEMENT