A sample of 55 weekly reports has been chosen from the population. Therefore, the sample size, n = 55.
Now, we have to find 90% and 95% confidence intervals for the population mean. A 90% confidence interval means that this interval will contain the population mean in 90% of the samples of the population, when taken repeatedly.
The formula to calculate the confidence intervals for the population mean is given as: [Ȳ - tα/2,n-1*(s/√n), Ȳ + tα/2,n-1*(s/√n)]
where Ȳ is the sample mean, s is the sample standard deviation and tα/2,n-1 is the value of the T-statistic at the value .
Why we use a T-statistic here is because we use Student's t distribution when estimating the mean of a normally distributed population in situations where the sample size is small and population standard deviation is unknown. You can use the t-table to find out the values at different values of .
a. To find 90% confidence interval, = 1-0.9 = 0.1
tα/2,n-1 = t0.025,54 = 1.67
Ȳ = 19.5
s= 5.9
The confidence interval will be given as : [19.5 - 1.67* (5.9/7.42), 19.5 + 1.67* (5.9/7.42)] = [18.17,20.83]
b. To find 95% confidence interval, = 1-0.95 = 0.05
tα/2,n-1 = t0.025,54 = 2
Ȳ = 19.5
s= 5.9
The confidence interval will be given as : [19.5 - 2* (5.9/7.42), 19.5 + 2* (5.9/7.42)] = [17.91,21.09]
Check My Work contacts made during the week. A sample of 55 weekly reports showed a...
Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 55 weekly reports showed a sample mean of 19.5 customer contacts per week. The sample standard deviation was 5.9. Provide 90% and 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel. 90% confidence interval, to 2 decimals: ( , ) 95% confidence interval, to 2 decimals: ( , )
Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 55 weekly reports showed a sample mean of 17.5 customer contacts per week. The sample standard deviation was 5.5. Provide 90% and 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel. 90% Confidence interval, to 2 decimals: ( , ) 95% Confidence interval, to 2 decimals: ( , )
Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 75 weekly reports showed a sample mean of 18.5 customer contacts per week. The sample standard deviation was 5.2. Provide 90% and 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel. a. 90% Confidence, to 2 decimals b. 95% Confidence, to 2 decimals
Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 75 weekly reports showed a sample mean of 18.5 customer contacts per week. The sample standard deviation was 5.1. Provide 90% and 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel. 90% Confidence, to 2 decimals: ( , ) 95% Confidence, to 2 decimals: ( , )
Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 85 weekly reports showed a sample mean of 17.5 customer contacts per week. The sample standard deviation was 5.8. Provide 90% and 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel. 90% confidence interval, to 2 decimals: 113 95% confidence interval, to 2 decimals
Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 75 weekly reports showed a sample mean of 17.5 customer contacts per week. The sample standard deviation was 5.6 . Provide 90% and 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel.
Check My Work eBook Video Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A s ample of 75 weekly reports showed a sample mean of 17. customer contacts per week. The sample standard deviation was SS. Provide 90% and 95% condence sales personnel. intervals for the population mean number of weekly customer contacts for the 90% confidence interval, to 2 decimals: 95% confidence interval, to 2 decimals:
Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 65 weekly reports showed a sample mean of 19.5 customer contacts per week. The sample standard deviation was 5.2. a)Develop a 90% confidence interval estimate of the population mean number of weekly customer contacts for the sales personnel. b)Develop a 95% confidence interval estimate of the population mean number of weekly customer contacts for the sales personnel. c)Discuss what happens to...
Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts mace during the week. A sample of 75 weekly reports showed a sampie mean of 18.5 customer contacts per week. The sample standard deviation was s.6. Provde 90% and 95% confidence intervals for the population mean nunter of weekly astomer conta tsfor the sales personnel. 90% Cerfidence interval, to 2 decimals: 95% confidence interval, to 2 deomas: A simple random sample of 70 items from a population with...
A simple random sample of 90 items from a population with = 9 resulted in a sample mean of 38. If required, round your answers to two decimal places. a. Provide a 90% confidence interval for the population mean. to b. Provide a 95% confidence interval for the population mean. to c. Provide a 99% confidence interval for the population mean. to Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample...