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 /n)

=1.965 * ( 2/ 450)

= 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 )


Related 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 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.
A random sample of 29 observations is used to estimate the population mean. The sample mean...
A random sample of 29 observations is used to estimate the population mean. The sample mean and the sample standard deviation are calculated as 130.2 and 29.60, respectively. Assume that the population is normally distributed Construct the 95% confidence interval for the population mean. Construct the 99% confidence interval for the population mean Use your answers to discuss the impact of the confidence level on the width of the interval. As the confidence level increases, the interval becomes wider. As...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT