One of the customer service department received the following number of calls during peak time over 12 time periods: 50, 49, 63, 54, 72, 76, 59, 66, 83, 64, 78, 61 It is known that the standard deviation of the number of calls during peak time is 16. Estimate the mean number of calls over peak time with a 98% confidence level.
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- In a random sample of 40 visitors to a certain theme park, it was determined that the mean amount of money spent per person at the park)including ticket price) was $93.43 per day with a standard deviation of $15. Construct and interpret a 99% confidence interval for the mean amount spent daily per person at the theme Park.The manager drew a sample of 30 delivery time in June and found that the mean delivery time is 17 minutes with standard deviation of 2.3 mins and a sample of 30 in August has mean and standard deviation of 17.5 minutes and 2 minutes. Do a hypothesis testing to determine if the time difference is significant or not. Use 95 per cent confidence levelThe board of a major credit card company requires that the mean wait time for customers when they call customer service is 3 minutes. To make sure that the mean wait time is not exceeding the requirement, an assistant manager tracks the wait times of 45 randomly selected calls. The mean wait time was calculated to be 3.4 minutes. Assuming the population standard deviation is 1.45 minutes, is there sufficient evidence that the mean wait time for customers is longer than 3 minutes with a 98% level of confidence? For the same problem as part A, construct an appropriate confidence interval based on the alpha level in A to support your conclusion from part A. Recall, if you rejected the null in part A, then your expected mean should not be included in the interval. If you failed to reject the null hypothesis, then the hypothesized mean should be in the interval. You should only need the appropriate critical value and margin of error in addition to values from part A. Correctly round your…
- The treasurer of a municipality claims that the average net worth of families living in this municipality is P730,000. A random sample of 50 families selected from this area produced a mean net worth of P860,000 with standard deviation of P65,000. Using 1% significance level, can we conclude that the claim is true? Also, find the 99% confidence interval of the true mean.The board of a major credit card company requires that the mean wait time for customers when they call customer service is at most 4.50 minutes. To make sure that the mean wait time is not exceeding the requirement, an assistant manager tracks the wait times of 52 randomly selected calls. The mean wait time was calculated to be 4.71 minutes. Assuming the population standard deviation is 0.82 minutes, is there sufficient evidence to say that the mean wait time for customers is longer than 4.50 minutes with a 98% level of confidence? Step 1 of 3 : State the null and alternative hypotheses for the test. Fill in the blank below.A major truck stop has kept extensive records on var-ious transactions with its customers. If a random sample of 18 of these records shows average sales of 63.84 gallonsof diesel fuel with a standard deviation of 2.75 gallons,construct a 99% confidence interval for the mean of thepopulation sampled.
- A former patient, who had both wisdom teeth extracted by two different dentists, claims that the average operating time (in minutes) performed by Dr. Reyes on all his patients is greater than the average operating time (in minutes) performed by Dr. Tupaz on all her patients. A sample of 102 patients attended to by Dr. Reyes had a mean operating time of 95.32 minutes with a population standard deviation of 29.27 minutes. On the other hand, 145 patients who went to the clinic of Dr. Tupaz gave a mean operating time of 84.11 minutes with a population standard deviation of 28.06 minutes. Assuming that both distributions are normal, is there sufficient evidence to support the former patient’s claim at alpha = 0.10? Use the p-value method.The board of a major credit card company requires that the mean wait time for customers when they call customer service is at most 3.00 minutes. To make sure that the mean wait time is not exceeding the requirement, an assistant manager tracks the wait times of 43 randomly selected calls. The mean wait time was calculated to be 3.28 minutes. Assuming the population standard deviation is 0.87 minutes, is there sufficient evidence to say that the mean wait time for customers is longer than 3.00 minutes with a 95% level of confidence? Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.A study conducted by the research department of a pharmaceutical company claims that the annual spending (per person) for prescription drugs for allergy relief, μ1 , is greater than or equal to the annual spending (per person) for non-prescription allergy relief medicine, μ2 . A health insurance company conducted an independent study and collected data from a random sample of 285 individuals for prescription allergy relief medicine. The sample mean is found to be 17.5 dollars/year, with a sample standard deviation of 4.8 dollars/year. They have also collected data for non-prescription allergy relief medicine. An independent random sample of 275 individuals yielded a sample mean of 18.5 dollars/year, and a sample standard deviation of 3.8 dollars/year. Since the sample size is quite large, it is assumed that the population standard deviation of the sales (per person) for prescription and non-prescription allergy relief medicine can be estimated by using the…
- A recent study compared the time spent together by single- and dual-earner couples. According to the records kept by the wives during the study, the mean amount of time spent together watching television among the single-earner couples was 61 minutes per day, with a standard deviation of 15.5 minutes. For the dual-earner couples, the mean number of minutes spent watching television was 48.4 minutes, with a standard deviation of 18.1 minutes. At the .01 significance level, can we conclude that the single-earner couples on average spend more time watching television together? There were 15 single-earner and 12 dual-earner couples studied.A study conducted by the research department of a pharmaceutical company claims that the annual spending (per person) for prescription drugs for allergy relief, μ1 , is greater than or equal to the annual spending (per person) for non-prescription allergy relief medicine, μ2 . A health insurance company conducted an independent study and collected data from a random sample of 225 individuals for prescription allergy relief medicine. The sample mean is found to be 17.6 dollars/year, with a sample standard deviation of 4.4 dollars/year. They have also collected data for non-prescription allergy relief medicine. An independent random sample of 220 individuals yielded a sample mean of 18.4 dollars/year, and a sample standard deviation of 3.8 dollars/year. Since the sample size is quite large, it is assumed that the population standard deviation of the sales (per person) for prescription and non-prescription allergy relief medicine can be estimated by using the…A recent survey conducted by the National Telecommunication Commission (NTC) aims to determine the average internet usage using smartphones (in minutes) per week. The researcher who surveyed 28 randomly selected people and found that the average time of internet usage using their smartphones per week is 140 minutes with a standard deviation of 30 minutes. Estimate a 95% confidence interval of the average time per minute
