Suppose a grocery store is considering the purchase of a new self-checkout machine that will get customers through the checkout line faster than their current machine. Before he spends the money on the equipment, he wants to know how mu faster the customers will check out compared to the current machine. The store manager recorded the checkout times, in seconds, for a randomly selected sample of checkouts from each machine. The summary statistics are provided in the tabl \[ \text { df }=90.71233 \] Compute the lower and upper limits of a95%confidence interval to estimate the difference of the mean checkout times fo customers. Estimate the difference for the old machine minus the new machine, so that a positive result reflects faster chec times with the new machine. Use the Satterthwaite approximate degrees of freedom, 90.71233 . Give your answers precise least three decimal places.
Question
Suppose a grocery store is considering the purchase of a new self-checkout machine that will get customers through the checkout line faster than their current machine. Before he spends the money on the equipment, he wants to know how mu faster the customers will check out compared to the current machine. The store manager recorded the checkout times, in seconds, for a randomly selected sample of checkouts from each machine. The summary statistics are provided in the tabl Compute the lower and upper limits of a95%confidence interval to estimate the difference of the mean checkout times fo customers. Estimate the difference for the old machine minus the new machine, so that a positive result reflects faster chec times with the new machine. Use the Satterthwaite approximate degrees of freedom, 90.71233 . Give your answers precise least three decimal places.
Solution
To compute the lower and upper limits of a 95% confidence interval to estimate the difference of the mean checkout times for customers, we will use the Satterthwaite approximate degrees of freedom, which is given as 90.71233.
Step 1: Calculate the standard error of the difference in means. The standard error can be calculated using the formula: SE = sqrt((s1^2 / n1) + (s2^2 / n2)) where s1 and s2 are the sample standard deviations, and n1 and n2 are the sample sizes.
Step 2: Calculate the t-value for a 95% confidence interval. The t-value can be calculated using the formula: t = (x1 - x2) / SE where x1 and x2 are the sample means.
Step 3: Calculate the margin of error. The margin of error can be calculated using the formula: ME = t * SE
Step 4: Calculate the lower and upper limits of the confidence interval. The lower limit can be calculated as: Lower limit = (x1 - x2) - ME The upper limit can be calculated as: Upper limit = (x1 - x2) + ME
Note: In this case, we want to estimate the difference for the old machine minus the new machine, so a positive result reflects faster checkout times with the new machine.
Make sure to use the provided summary statistics from the table to calculate the necessary values accurately.
Similar Questions
The lower limit of the 95% confidence interval for the mean of the differences is 0.241.Which ONE of the following is the correct way to calculate the value of the upper limit?0.539 + 2 x 0.2980.539 + 2 x 0.980.539 + 0.2980.539 + 0.241
The waiting time to check out of a supermarket has had a population mean of 11.02 minutes.Recently,in an effort to reduce the waiting time,the supermarket has experimented with a system in which there is a single waiting line with multiple checkout servers.A sample of 50 customers was selected,and their mean waiting time to check out was 9.79 minutes with a sample standard deviation of 5.6 minutes.Complete parts(a)through(d). Click here to view page 1 of the critical values for the t distribution. Click here to view page 2 of the critical values for the t Distribution. a.At the 0.01 level of significance,using the critical value approach to hypothesis testing,is there evidence that the population mean waiting time to check out is less than 11.02 minutes? What are the null and alternative hypotheses for this test?
The waiting time to check out of a supermarket has had a population mean of 8.49 minutes.Recently,in an effort to reduce the waiting time,the supermarket has experimented with a system in which there is a single waiting line with multiple checkout servers.A sample of 90 customers was selected,and their mean waiting time to check out was 7.85 minutes with a sample standard deviation of 4.9 minutes.Complete parts(a)through(d) H0:μ≥8.49 H1:μ<8.49 What is the test statistic for this test? -1.2391 (Round to four decimal places as needed.) What is the critical value for this test?
A researcher wants to estimate the proportion of depressed individuals taking a new anti-depressant drug who find relief. A random sample of 275 individuals who had been taking the drug is questioned; 214 of them found relief from depression. Based upon this, compute a 95% confidence interval for the proportion of all depressed individuals taking the drug who find relief. Then find the lower limit and upper limit of the 95% confidence interval.Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places. (If necessary, consult a list of formulas.)Lower limit: Upper limit:
The table below shows a sample of 4 commuters in Chicago showed the the commuting times in minutes. Find the following:time33.936.937.927.5At 98% confidence level, the maximum error of estimate (positive) is equal to: Find the 98% lower confidence interval of the true mean commuting times.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.