Consider a standardized two-sided CUSUM with k = 0.2 and h = 8. (a) Use Siegmund's procedure to evaluate the in-control ARL performance of this scheme. (b) Find the out-of-control ARL for 8 = 0.5
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- amu mun nuo 19. Commute Time to Work A survey of 15 large U.S. cities finds that the average commute time one way is 25.4 minutes. A chamber of commerce executive Technology Step by Step TI-84 Plus Hypothesis Test for the M 1. Enter the data values into L1. 2. Press STAT and move the c 3. Press 2 for T-Test. Step by Step 4. Move the cursor to Data and 5. Type in the appropriate valuThe data in the contingency table to the right count the number of male and female shoppers who accept or reject a discounted offer in return for supplying a retailer with an email address. Consider using these data to test the null hypothesis that gender and acceptance are independent (H0). Male Female Accept 105 95 Reject 95 105 (c) Find the value of x2(chi-square) for testing H0. The value of x2 (chi-square)for testing H0 is _____. (Round to three decimal places as needed.) (d) Find the p-value for testing H 0. (Round to three decimal places as needed.)What is the meaning (eq) of the formula L in the question (f)?
- Is sunscreen with an SPF 30 rating more effective at preventing sunburn than sunscreen with an SPF 15 rating? At the a = 0.01 level of significance, test the claim that SPF 30 sunscreen is better at preventing sunburn than SPF 15 sunscreen. Let pi5 represent the proportion of SPF 15 sunscreen wearers who get a sunburn and p30 represent the proportion of SPF 30 sunscreen wearers who get a sunburn.(Round your results to three decimal places) Which would be correct hypotheses for this test? O Ho:P15 p30 O Ho: P15 = P30, H1: P15 > P30 O Ho:P15 = p30, H1: P15 7 P30 O Ho:P15 = P30, H1:P15 < P30 A random sample of 100 beach-goers were given SPF 15 sunscreen and asked to lay in the sun for 1 hour. 16 of them later reported a sunburn. Another random sample of 100 beach-goers were given SPF 30 sunscreen and asked to lay in the sun for 1 hour. 13 of them later reported a sunburn. Find the test statistic: Give the P-value: Which is the correct result: O Do not Reject the Null Hypothesis O Reject…From a CoR test, the r-value comes as 2.30. What is wrong with this value?For a hypothesis test of the claim that the mean amount of sleep for adults is less than 6 hours, technology output shows that the hypothesis test has power of 0.4733 of supporting the claim that μ<6 hours of sleep when the actual population mean is 4.0 hours of sleep. Interpret this value of the power, then identify the value of β and interpret that value.
- A television manufacturer claims that (at least) 90% of its TV sets will not need service during the first 3 years of operation. A consumer agency wishes to check this claim, so it obtains a random sample 100 purchasers and asks each whether the set purchased needed repair during the first 3 years after purchase. Let p be the sample proportion of responses indicating no repair (so that no repair of n = is identified with a success). Let p denote the actual proportion of successes for all sets made by this manufacturer. The agency does not want to claim false advertising unless sample evidence strongly suggests that p < 0.9. The appropriate hypotheses are then Ho: p = 0.9 versus H: p < 0.9. %3Dnts ervices An educational psychologist is examining response times to an on-screen stimulus. The researcher believes there might be a weak effect from age, but expects a more pronounced effect for different color contrasts. She decides to examine a black on white (B/W) combination compared to 2 milliseconds): alternatives: red on white (R/W) and yellow on blue (Y/B). Here is the data for response times (in Source Age (A) Color (B) Interaction (A x B) Error SS 759.083 Os 604.333 o 373.333 5559.25 Age OF 16-17 18-19 20-21 df 2 2 4 B/W 25 9 21 Statistical software was used to run a two-way Anova test. 63 21 10 30 15 AM 33 25 14 16 ANA SAANNNNN 17 12 20 40 17 19 25 24 22 28 00 Color Scheme MS 379.542 of 302.167 OF 93.333 OF 88.242 OF R/W Y/B 20 19 16 43 20 31 20 36 37 35 ២២១១ 19 39 26 29 20 31 41 19 AMATENEG 19 33 22 43 28 36 43 26 F-ratio 4.301 $ 0° 3.424 O 1.058 P-value 0.0177 O 0.0387 OF 0.3849 Σ ||₁PLEASE SHOW THE EXCEL OUTPUT ONLY OF THESE. 1. There is no significant difference between the perceived level of riskiness of thrill behavior in males and in females. (Use α= 0.05) 2. There is no significant difference between the perceived level of riskiness of rebellious behavior in males and in females. (Use α= 0.10) 3.There is no significant difference between the level of riskiness of thrill-seeking items and rebellious items as perceived by the males. (Use α= 0.05).
- = 1. The height of male giraffes follows a normal distribution with µ 18 feet and o= 0.9 feet². (a) What is the probability that a male giraffe has a height between 19 and 22 feet? Write out the solution in integral form and use the following R code to compute the numerical answer: pnorm (22, 18,0.9) - pnorm(19, 18,0.9). (b) What height would a male giraffe need to be in order to be in the top 10 percent? Write out the equation we would need to solve in order to answer this question, then use the qnorm() function in R to find this score. (c) What is the probability that a male giraffe will grow shorter than 16 feet. (d) The tallest male giraffe in the world is measured to be 21 feet according to the Guinness World Records. What is the probability that a giraffe would grow higher than 21 feet.A manufacturer wants to increase the shelf life of a line of cake mixes. Past records indicate that the average shelf life of the mix is 231 days. After a revised mix has been developed, a sample of nine boxes of cake mix gave these shelf lives (in days): 230,233, 235, 236, 233, 240, 240, 241, and 237. Using α = 0.025, has the shelf life of the cake mix increased? Multiple Choice Yes, because computed t is greater than the critical value. Yes, because computed t is less than the critical value. No, because computed t lies in the region of acceptance. No, because 236.11 is quite close to 231.What percentage of scores is between a z of ‐2 and a z of +2? Why is this important?
