Question

In: Statistics and Probability

A random sample of 25 employees for the retailer showed a sample mean of 15.1 minutes...

A random sample of 25 employees for the retailer showed a sample mean of 15.1 minutes and a standard deviation of 3 minutes. Assume that the time spent by employees on personal phone calls is normally distributed. Let μ denote the mean time spent by employees spent on personal phone calls.

(a) An employee group for a national retailer claims that the mean time spent by employees on personal phone calls is more than 20 minutes per day. Specify the correct null and alternative hypotheses to investigate the employee group’s claim.

(b) Find the value of the standardized test statistic.

(c) Find the rejection region at α = .05 and state your conclusion in the context of the problem.

(d) State the Type I error based on the context of the problem

Solutions

Expert Solution

n=25,  = 20,   = 0.05

= 15.1 , s= 3

a)

null and alternative hypothesis is

Ho:     20

H1:   > 20

b)

formula for test statistics is

t = -8.167

test statistics= -8.167

c)

calculate t critical value for right tailed test with df= n-1 = 25-1 = 24

using t table we get critical value as follows

Critical value= 1.711

rejection region is as follows,

decision rule is

Reject Ho if ( t ) > ( 1.711)

here, ( t= -8.167 ) < ( 1.711 )

Hence, we can say,

Null hypothesis is NOT rejected.

There is NOT sufficient evidence to support th claim that the mean time spent by employees on personal phone calls is more than 20 minutes per day.

d)

type I error occurs when we reject the true null hypothesis.

In these context type I error occurs when we conclude that the mean time spent by employees on personal phone calls is more than 20 minutes per day.

but in fact  mean time spent by employees on personal phone calls is less than or equal to 20 minutes per day,


Related Solutions

The records of a random sample of 25 Amazon employees in a large city showed that...
The records of a random sample of 25 Amazon employees in a large city showed that the average years these employees had worked for the Amazon was ?̅= 4 years. Assume that we know that the population distribution of years Amazon employees have spent with the company is approximately Normal, with standard deviation ? = 1.3 years. Assume all conditions have been met. Construct and interpret a 99% confidence interval for the true mean years the population of Amazon employees...
A random sample of 25 employees of a local utility firm showed that their monthly incomes...
A random sample of 25 employees of a local utility firm showed that their monthly incomes had a sample standard deviation of $112. Provide a 90% confidence interval estimate for the standard deviation of the incomes for all the firm's employees.
A random sample of 25 workers showed that the average workweek in the United States is...
A random sample of 25 workers showed that the average workweek in the United States is down to only 35 hours, largely because of a rise in part-time workers. The variance of the sample was 18.49 hours. Assuming that the population is normally distributed, calculate a 98% confidence interval for the population standard deviation.
A random sample of 20 purchases showed the amounts in the table​ (in $). The mean...
A random sample of 20 purchases showed the amounts in the table​ (in $). The mean is ​$50.76 and the standard deviation is ​$19.48 ​a) Construct a 80​% confidence interval for the mean purchases of all​ customers, assuming that the assumptions and conditions for the confidence interval have been met. ​b) How large is the margin of​ error? ​c) How would the confidence interval change if you had assumed that the standard deviation was known to be $20? 60.91 64.3...
A random sample of 20 purchases showed the amounts in the table (in $). The mean...
A random sample of 20 purchases showed the amounts in the table (in $). The mean is $49.88 and the standard deviation is $20.86. B) What is the margin of error ?A) Construct a 80% confidence interval for the mean purchases of all customers, assuming that assumptions and conditions for confidence interval have been met. (What is the confidence interval) ? C) How would the confidence interval change if you had assumed that the standard deviation was known to be...
A random sample of 2020 purchases showed the amounts in the table​ (in $). The mean...
A random sample of 2020 purchases showed the amounts in the table​ (in $). The mean is ​$49.8149.81 and the standard deviation is ​$21.9521.95. ​a) Construct a 8080​% confidence interval for the mean purchases of all​ customers, assuming that the assumptions and conditions for the confidence interval have been met. ​b) How large is the margin of​ error? ​c) How would the confidence interval change if you had assumed that the standard deviation was known to be ​$2222​? A) The...
A random sample of 20 purchases showed the amounts in the table​ (in $). The mean...
A random sample of 20 purchases showed the amounts in the table​ (in $). The mean is ​$49.71 and the standard deviation is ​$20.12 ​a) Construct a 99​% confidence interval for the mean purchases of all​ customers, assuming that the assumptions and conditions for the confidence interval have been met. ​b) How large is the margin of​ error? ​c) How would the confidence interval change if you had assumed that the standard deviation was known to be ​$21? 40.97 68.94...
A random sample of 16 pharmacy customers showed the waiting times below (in minutes). Also, the...
A random sample of 16 pharmacy customers showed the waiting times below (in minutes). Also, the sample is from a normal population. Note that σ is unknown. 21 22 22 17 21 17 23 20 20 24 9 22 16 21 22 21 The sample mean is 19.875 and the sample standard deviation is 3.65. Which of the following represents the 80 percent confidence interval for µ? Select one: a. [13.75, 25.25] b. [18.65, 21.10] c. [19.55, 20.425] d. [18.8,...
A local retailer claims that the mean waiting time is less than 8 minutes. A random...
A local retailer claims that the mean waiting time is less than 8 minutes. A random sample of 20 waiting times has a mean of 6.8 minutes with a standard deviation of 2.1 minutes. At a = 0.01, test the retailer's claim. Assume the distribution is normally distributed. Round the test statistic to the nearest thousandth.
A local retailer claims that the mean waiting time is less than 8 minutes. A random...
A local retailer claims that the mean waiting time is less than 8 minutes. A random sample of 20 waiting times has a mean of 6.3 minutes with a standard deviation of 2.1 minutes. At α = 0.01, test the retailerʹs claim. Assume the distribution is normally distributed. Use any method, however, follow the PHANTOMS acronym. P - Parameter Statement H - Hypotheses A - Assumptions & Conditions N - Name the Test and state the curve you're using T...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT