Construct a 94% confidence interval for the difference between the mean lifetimes of two kinds of light bulbs, given that a random sample of 40 light bulbs of the first kind lasted on the average 44 hours of continuous use and 50 light bulbs of the second kind lasted on the average 40 hours of continuous use. The population standard deviations are known to be oj = 26 and o2 = 22. %3D
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- Construct a 95% confidence interval of the population standard deviation for the 18. The thickness of a plastic film (in mm) on a substrate material is thought to be influenced by the temperature at which the coating is applied. In completely randomized experiment, 11 substrates are coated at 52°C, resulting in a sample mean coating thickness of 2.63 mm and a sample standard deviation of 0.26 mm. Another 13 substrates are coated at 66°C for which a sample mean of 2.53 mm and a sample standard deviation of 0.51 mm are observed. The Minitab output of the test for two variances is as given. Assume that the first and second samples given in the output are the substrates coated at 52°C and 66°C respectively. Assume also that the distribution of the thickness of plastic films is normal. Test and CI for Two Variances Method Null hypothesis Alternative hypothesis o (First)/o (Second) = 1 o (First)/o( Second) # 1 F method was used. This method is accurate for normal data only Statistics Sample…A person read that the average number of hours an adult sleeps on Friday night to Saturday morning was 7.4 hours. The researcher feels that college students sleep less than 7.4 hours on average. The researcher randomly selected 16 students and found that on average they slept 6.2 hours. The standard deviation of the sample is 1.1 hours. At α=0.05, is there enough evidence to say that college students sleep less than 7.4 hours on average? Assume that the population is approximately normally distributed. Use the P-value method and tables. Compute the t test value. What is the correct interval for the P-value? Does this reject the hypothesis? Thank you very much!For a certain knee surgery, a mean recovery time of 13 weeks is typical. With a new style of physical therapy, a researcher claims that the mean recovery time, μ, is less than 13 weeks. In a random sample of 32 knee surgery patients who practiced this new physical therapy, the mean recovery time is 12.8 weeks. Assume that the population standard deviation of recovery times is known to be 1.1 weeks. Is there enough evidence to support the claim that the mean recovery time of patients who practice the new style of physical therapy is less than 13 weeks? Perform a hypothesis test, using the 0.10 level of significance. (a) State the null hypothesis Ho and the alternative hypothesis H₁. H Ho: O H₁:0 OO 020 ローロ OO ? (b) Perform a Z-test and find the p-value. Here is some information to help you with your Z-test. • The value of the test statistic is given by x-μ √n • The p-value is the area under the curve to the left of the value of the test statistic. Standard Normal Distribution 04 Step 1:…
- An engineer working for a tire manufacturer investigates the average life of a new material to be used in the manufacturing process. For this purpose, he builds a sample of 13 tires and tests them in a laboratory until the end of their useful life is reached. The data obtained from the sample are as follows: average of 60897.9 km and standard deviation of 2734.8 km. The engineer would like to demonstrate that the average service life of the tire with the new material exceeds 60,000 km. a) State the null and alternative hypothesis to be stated in this experiment. Test the stated hypotheses using an α = 0.05 What conclusions are reached? b) Find a 95% confidence interval for the average tire life with the new material. Interpret your answer. (RESPOND MANUALLY) (RESPOND MANUALLY)Bowker and Lieberman (1972) gave the results for tests of two thermostats used in irons that were made by an old supplier and a new supplier. 18 thermostats were sampled from the old supplier and 21 thermostats were sampled from the new supplier. The actual temperatures measured with a thermocouple and were recorded. The sample from the old supplier had a mean of x₁ =549.8 and a standard deviation of s₁=9.6. The sample from the new supplier had a mean of x₂=550.4 and a standard deviation of s2=11.2. Determine the value of the pooled standard deviation Sp Sp = ... (Record to two decimal places.) 4Do male and female servers at Swank Bar work the same number of hours? A sample of 65 female servers worked an average of 37 hours per week, with a standard deviation of 2. A sample of 65 male servers worked an average of 22 hours per week, with a standard deviation of 3.Let μ1μ1 and μ2μ2 represent the typical number of hours worked by all female and male servers at Swank Bar, respectively. Estimate with a 85% confidence level how many more hours female servers work. Round answers to the nearest hundredth.< μ1−μ2μ1-μ2 <Which of the following does your data suggest? Female servers work more hours Male servers work more hours Female and male servers work about the same number of hours
- The sociologist finds that for a certain population, the mean number of years of educations is 13.2 years witha standard deviation of 3.04 years. From a certain region, a random sample of 62 people is drawn from thispopulation and recorded a sample mean of 13.96 years. Test the claim that the mean number of years ofeducation is the same from the population with 0.05 level of significance.The lifetime of a certain brand of battery is known to have a standard deviation of 19.8 hours. Suppose that a random sample of 80 such batteries has a mean lifetime of 35.8 hours. Based on this sample, find a 95% confidence interval for the true mean lifetime of all batteries of this brand. Then give its lower limit and upper limit.The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $14,124. The following year, a random sample of 20 two-year institutions had a mean of $17,124 and a standard deviation of $3700. Is there sufficient evidence at the alpha level of 0.05 to conclude that the mean cost has increased. Show 6 steps of hypothesis
- A researcher claims that the average life span of mice can be extended when the calories in their diet are reduced by approximately 40% from the time they are weaned. The restricted diets are enriched to normal levels by vitamins and protein. Suppose that a sample of 10 mice is fed a normal diet and has anaverage life span of 32.1 months with a standard deviation of 3.2 months, while a sample of 15 mice is fed the restricted diet and has an average life span of 37.6 months with a standard deviation of 2.8 months. Assume the distributions of life spans for the regular and restricted diets are approximately normal withequal variance.(a) State the appropriate hypotheses to test the researcher’s claim.(b) Conduct the test at the 0.05 level of significance. What do you conclude?2) The manufacturer of a certain brand of automobiles component claims that mean life of these component is 70 months. An independent agency wants to check this claim and took a random sample of 40 components and found that the mean life for this sample is 67 months with a standard deviation of 5 months. At 5% LOS, can we conclude that the mean life of these component is less than 70 months?A recent study stated that if a person chewed gum, the average number of sticks of gum he or she chewed daily was 8. To test the claim, the researcher selected a random sample of 36 gum chewers and found the mean number of sticks of gum chewed per day was 8. The standard deviation of population is 1. at a = 0. 05, is the number of sticks of gum a person chews per day actually greater than 8. There is not enough evidence to support the claim that average is greater than 8. There is enough evidence to support the claim that average is greater than 8. There is enough evidence to support the claim that average is smaller than 8. There is not enough evidence to support the claim that average is greater than 8.
