The customers at a bank complained about long lines and the time they had to spend waiting for service. It is known that the customers at this bank had to wait 11 minutes, on average, before being served. The management made some changes to reduce the waiting time for its customers. A sample of 61 customers taken after these changes were made produced a
Use the p-value approach. Use the t distribution table to find a
Enter your answer; the range for the p-value, lower bound< p-value <Enter your answer; the range for the p-value, upper bound
Use the critical-value approach. Round your answers to three decimal places.
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- It currently takes users a mean of 12 minutes to install the most popular computer program made by RodeTech, a software design company. After changes have been made to the program, the company executives want to know if the new mean is now different from 12 minutes so that they can change their advertising accordingly. A simple random sample of 91 new customers are asked to time how long it takes for them to install the software. The sample mean is 11.7 minutes with a standard deviation of 1.6 minutes. Perform a hypothesis test at the 0.05 level of significance to see if the mean installation time has changed. Step 2 of 3 : Compute the value of the test statistic. Round your answer to three decimal places.arrow_forwardAccording to a news program, Americans take an average of 4.9 days off per year because of illness. The manager of a large chain of grocery stores wants to know if the employees at the grocery store, on average, take fewer days off than the national average. The manager selects a random sample of 80 employees in the company and found the sample mean number of days off for the 80 employees was 4.75 days with a standard deviation of 0.9 days. When the manager performed a significance test, the P-value was 0.07. What is the meaning of this P-value in context?arrow_forwardFran is training for her first marathon, and she wants to know if there is a significant difference between the mean number of miles run each week by group runners and individual runners who are training for marathons. She interviews 42 randomly selected people who train in groups and finds that they run a mean of 47.1 miles per week. Assume that the population standard deviation for group runners is known to be 4.4 miles per week. She also interviews a random sample of 47 people who train on their own and finds that they run a mean of 48.5 miles per week. Assume that the population standard deviation for people who run by themselves is 1.8 miles per week. Test the claim at the 0.01 level of significance. Let group runners training for marathons be Population 1 and let individual runners training for marathons be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.arrow_forward
- Many people consider their smart phone to be essential! Communication, news, Internet, entertainment, photos, and just keeping current are all conveniently possible with a smart phone. However, the battery better be charged or the phone is useless. Battery life of course depends on the frequency, duration, and type of use. One study involving heavy use of the phones showed the mean of the battery life to be 11.25 hours with a standard deviation of 3.3 hours. Then the battery needs to be recharged. Assume the battery life between charges is normally distributed. (a) Find the probability that with heavy use, the battery life exceeds 12 hours. (Round your answer to four decimal places.)(b) You are planning your recharging schedule so that the probability your phone will die is no more than 5%. After how many hours should you plan to recharge your phone? (Round your answer to the nearest tenth of an hour.) hoursarrow_forwardA new small business wants to know if its current radio advertising is effective. The owners decide to look at the mean number of customers who make a purchase in the store on days immediately following days when the radio ads are played as compared to the mean for those days following days when no radio advertisements are played. They found that for 1010 days following no advertisements, the mean was 18.618.6 purchasing customers with a standard deviation of 1.51.5 customers. On 77 days following advertising, the mean was 20.320.3 purchasing customers with a standard deviation of 0.90.9 customers. Test the claim, at the 0.020.02 level, that the mean number of customers who make a purchase in the store is lower for days following no advertising compared to days following advertising. Assume that both populations are approximately normal and that the population variances are equal. Let days following no advertisements be Population 1 and let days following advertising be Population 2.…arrow_forwardA new small business wants to know if its current radio advertising is effective. The owners decide to look at the mean number of customers who make a purchase in the store on days immediately following days when the radio ads are played as compared to the mean for those days following days when no radio advertisements are played. They found that for 10 days following no advertisements, the mean was 18.3 purchasing customers with a standard deviation of 1.8 customers. On 7 days following advertising, the mean was 19.4 purchasing customers with a standard deviation of 1.6 customers. Test the claim, at the 0.02 level, that the mean number of customers who make a purchase in the store is lower for days following no advertising compared to days following advertising. Assume that both populations are approximately normal and that the population variances are equal. Let days following no advertisements be Population 1 and let days following advertising be Population 2. Step 3 of 3: Draw a…arrow_forward
- A pharmaceutical company needs to know if its new cholesterol drug, Praxor, is effective at lowering cholesterol levels. It believes that people who take Praxor will average a greater decrease in cholesterol level than people taking a placebo. After the experiment is complete, the researchers find that the 44 participants in the treatment group lowered their cholesterol levels by a mean of 18.7 points with a standard deviation of 3.3 points. The 39 participants in the control group lowered their cholesterol levels by a mean of 18.1 points with a standard deviation of 2.1 points. Assume that the population variances are not equal and test the company’s claim at the 0.02 level. Let the treatment group be Population 1 and let the control group be Population 2. Step 1 of 3 : State the null and alternative hypotheses for the test. Fill in the blank below. H0: μ1=μ2 Ha : μ1⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯μ2 Step 2 of 3: what is the test statistic Step 3 of 3: draw a conclusion, fail or reject. Is…arrow_forwardIt seems these days that college graduates who are employed full time work more than 40 hour weeks. Data are available that can help us decide if this is true. A survey was recently sent to a group of adults selected at random. There were 22 respondents who were college graduates employed full time. The mean number of hours worked per week by these 22 respondents was 43 hours, with a standard deviation of 8 hours. Assume that the population of hours worked per week by college graduates employed full time is normally distributed with mean. Can we conclude that mean is greater than 40 hours? Use the 0.1 level of significance. Perform a one tailed test, and complete the parts below.arrow_forwardIt currently takes users a mean of 26 minutes to install the most popular computer program made by RodeTech, a software design company. After changes have been made to the program, the company executives want to know if the new mean is now different from 26 minutes so that they can change their advertising accordingly. A simple random sample of 37 new customers are asked to time how long it takes for them to install the software. The sample mean is 27.9 minutes with a standard deviation of 7.7 minutes. Perform a hypothesis test at the 0.05 level of significance to see if the mean installation time has changed. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.arrow_forward
- A pharmaceutical company needs to know if its new cholesterol drug, Praxor, is effective at lowering cholesterol levels. It believes that people who take Praxor will average a greater decrease in cholesterol level than people taking a placebo. After the experiment is complete, the researchers find that the 34 participants in the treatment group lowered their cholesterol levels by a mean of 22.2 points with a standard deviation of 3.4 points. The 42 participants in the control group lowered their cholesterol levels by a mean of 21.2 points with a standard deviation of 1.8 points. Assume that the population variances are not equal and test the company's claim at the 0.10 level. Let the treatment group be Population 1 and let the control group be Population 2. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. Ho:₁ = ₂ Ha: M •M₂arrow_forwardA new small business wants to know if its current radio advertising is effective. The owners decide to look at the mean number of customers who make a purchase in the store on days immediately following days when the radio ads are played as compared to the mean for those days following days when no radio advertisements are played. They found that for 13 days following no advertisements, the mean was 23.9 purchasing customers with a standard deviation of 1.9 customers. On 6 days following advertising, the mean was 24.7 purchasing customers with a standard deviation of 1.6 customers. Test the claim, at the 0.01 level, that the mean number of customers who make a purchase in the store is lower for days following no advertising compared to days following advertising. Assume that both populations are approximately normal and that the population variances are equal. Let days following no advertisements be Population 1 and let days following advertising be Population 2. Step 3 of 3: Draw a…arrow_forwardResearchers wants to determine if low-fat diets or low-carb diets are more effective for weight loss. Suppose 100 people were placed on a low-fat diet and at the end of 3 months the mean weight loss was 10.5 lbs witha standard deviation of 12.5 lbs. There were another 100 people placed on a low-carb diet and at the end of 3 months the mean weight loss was 16.5 lbs with a standard deviation of 10.2 lbs. Does mean weight lossdiffer between the two diets? What are the hypotheses??arrow_forward
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