Construct a 99 % 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 41 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 o1 = 26 and o2 = 22. Solution For a = 0.06, we find from Table III that z0.03 = 94% confidence interval for µi – 1.88. Therefore, the I uz is 262 (418 - 402) – 1.88 · V 40. 222 < H1 - 42 50 262 222 + 40 < (418 – 402) +1.88 · 50
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- 3.3.15 Use z scores to compare the given values. Based on sample data, newborn males have weights with a mean of 3293.7 g and a standard deviation of 551.3 g. Newborn females have weights with a mean of 3000.6 g and a standard deviation of 590.5 g Who has the weight that is more extreme relative to the group from which they came; a male who weighs 1700 g or a female who weighs 1700 g? V has the weight that is more extreme. Since the z score for the male is z= and the z score for the female isz= the (Round to two decimal places.)A manufacturer bonds a plastic coating to a metal surface. A random sample of nine observations on the thickness of this coating is taken from a week's output and the thickness (in millimeters) of these observations are shown below. Assuming normality, find a 90 % confidence interval for the population variance. 19.7 20.5 Click the icon to view a table of lower critical values for the chi-square distribution. Click the icon to view a table of upper critical values for the chi-square distribution. 21.8 18.2 Find the 90% confidence interval. 0<²0 (Round to four decimal places as needed.) 21.6 SCIOD 19.3 19.7 20.2 20.7A sample is selected from a normal population with a mean of µ = 40. If the sample mean is M = 45, which of the following combinations would make the sample mean an extreme, unrepresentative value for the population
- A pharmaceutical company has developed a drug that is expected to reduce hunger. To test the drug, three samples of rats are selected with n=10n=10 in each sample. The first sample receives the drug every day. The second sample is given the drug once a week, and the third sample receives no drug at all (the control group). The dependent variables is the amount of food eaten by each rat over a 1-month period. These data are analyzed by an ANOVA, and the results are reported in the following summary table. Fill in all missing values in the table. (Hint: Start with the df column.) S.S. d.f. M.S. F Between 6.68 Within 4.35 TOTAL Use the =FDIST() function in Excel to locate the p-value for this ANOVA:p-value = Report p-value accurate to at least 6 decimal places.If you use a significance level of α=.05α=.05, what would you conclude about these treatments?The level of nitrogen oxides (NOX) and nonmethane organic gas (NMOG) in thedusLove uo useful life (150,000 miles of driving) of cars of a particular model varies Normall/ih nean 80 mg/mi and standard deviation 4 mg/mi. A company has 25 cars of this model in its et find the level L such that the probability that the average NOX+NMOG level 7 for the fleet is greater than L is only 0.01. Give your answer to three decimal places.A certain IQ test is known to have a population mean of 100 and standard deviation of 15 in the general population. You want to test whether psychology majors have a different average IQ than the population as a whole. Assume the variance of IQ is the same for Psych majors as it is in the general population. Suppose that Psychology majors actually have an average IQ of 108. If you do a 2-tailed test at α= .05 with a sample of 56 Psychology majors, you will be able to reject the null hypothesis if the mean IQ of your sample is below [L] or above [H]. Find L and H values. Options listed below. [L] answer choices: 96.08, 98.00, 103.92, 104.08, 110.00, 111.92. [H] Answer choices: 96.08, 98.00, 103.92, 104.08, 110.00, 111.92.
- A sample of n = 16 scores is selected from a population with μ = 80 with σ = 20. On average, how much error would be expected between the sample mean and the population mean?For the general population, the average fatigue level is μ = 50. A researcher is testing the hypothesis that engaging a meditation exercise reduces employee fatigue. A sample of n = 30 participants is obtained, and each participant engaged in a meditation exercise for 20 minutes at work each workday for a month. This sample’s average fatigue level was M = 46 with a sample variance of 100 after the month. Please write out the four steps for this hypothesis test assuming an alpha level of .05. Step 1: State your hypotheses Step 2: Find your critical regions in a distribution Step 3: Data Analysis Step 4: Making a decisionRecent results suggest that children with ADHD tend to watch more TV than children who are not diagnosed with the disorder. To examine this relationship, a researcher obtains a random sample of n = 36 children, 8 to 12 years old, who have been diagnosed with ADHD. Each child is asked to keep a journal recording how much time each day is spent watching TV. The average daily time for the sample is M = 4.9 hours. It is known that the average time for the general population of 8 to 12-year-old children is = 4.1 hours with = 1.8. (Hint: Be sure to use the correct test statistic). Are the data sufficient to conclude that children with ADHD watch significantly more TV than children without the disorder? Use a two-tailed test with = .05. (Be sure to give your conclusion, too) If the researcher had used a sample of n = 9 children and obtained the same sample mean, would the results be sufficient to reject H0? (Be sure to give your conclusion, too) Compute Cohen's d for this study. What is…
- A random sample of n = 19 winter days in Denver gave a sample mean pollution index x1 = 43. Previous studies show that o1 = 10. For Englewood (a suburb of Denver), a random sample of n2 = 18 winter days gave a sample mean pollution index of x2 = 34. Previous studies show that o2 = 13. Assume the pollution index is normally distributed in both Englewood and Denver. Do these data indicate that the mean population pollution index of Englewood is different (either way) from that of Denver in the winter? Use a 1% level of significance. (a) What is the level of significance? State the null and alternate hypotheses. O Ho: H1 H2 O Ho: H1 = l2; H1: H1 < µ2 (b) What sampling distribution will you use? What assumptions are you making? O The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations. O The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations. O The Student's…Consider the least-squares regression line Cognition = 85.3 – 2.54 Years %3D that predicts the score on a cognition test given the number of years playing hockey. Find the predicted cognition score for the given case. Two intervals are shown below: one is a 95% confidence interval for the mean response and the other is a 95% prediction interval for the response. Which is the 95% confidence interval for the mean response? Given a person who has played hockey for 5 years: Interval I: (28.2, 117.0) and Interval lI: (62.2, 83.0) O The predicted cognitive score is 72.60 for a person who has played hockey for 5 years. The 95% confidence interval for the mean response is ( 28.2, 117.0). O The predicted cognitive score is 72.60 for a person who has played hockey for 5 years. The 95% confidence interval for the mean response is (62.2, 83.0). O The predicted cognitive score is 85.3 for a person who has played hockey for 5 years. The 95% confidence interval for the mean response is ( 28.2,…A nationwide survey in 2001 revealed that U.S. grade-school children spend an average of µ = 8.0 hours per week doing homework. The distribution is normal with σ = 2.5. Last year, a sample of n = 100 grade-school children was given the same survey. For this sample, the mean number of homework hours was 7.4. Has there been a significant change in the homework habits of grade-school children? Test with α = .05. For each problem students will write out all steps of hypothesis testing: State the null and research hypotheses. Set the level of risk associated with the null hypothesis. Select the appropriate test statistic. Compute the test statistic value (called the obtained value). Determine the value needed for rejection of the null hypothesis using the appropriate table of critical values for the particular statistic. Compare the obtained value and the critical value. Make your decision