Procedure 1. Control solution A was analyzed daily, and the following values were obtained: Days of Running 1 2345678 9 10 11 12 13 14 15 16 17 18 19 20 2. Compute for the following: a. Mean b. Standard Deviation c. ± 1 SD d. ± 2 SD Control Values (mg/dL) 118 120 121 119 125 118 122 116 124 123 117 117 121 120 120 119 121 123 120 122 e. ± 3 SD f. Coefficient of Variation g. Variance
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- Find the mean hourly cost when the cell phone described above is used for 240 minutes.A large automobile insurance company selected samples of single and married male policyholders and recorded the number who made an insurance claim over the preceding three-year period. Single Policyholders Married Policyholders ni = 412 n2 = 778 Number making claims = 75 Number making claims = 111 a. Use a= 0.05. Test to determine whether the claim rates differ between single and married male policyholders. z-value (to 2 decimals) p-value (to 4 decimals) We cannot conclude that there is the difference between claim rates. b. Provide a 95% confidence interval (to 4 decimals) for the difference between the proportions for the two populations. Enter negative answer as negative number.Dr. Johnson is interested in the difference in mean systolic blood pressure between groups. He performs analysis and finds a mean difference of 0.0005 mmHg in systolic blood pressure with a p value of 0.001. Dr. Johnson's findings are: a) statistically insignificant and practically significant b) statistically significant and practically significant c) statistically insignificant and practically insignificant d) statistically significant but practically ins
- The mean arterial pressure (or MAP, average arterial pressure in one cardiac cycle) is a measure for perfusion status. You are interested to know if it is possible to predict an adult's MAP (Y) based on the body surface area (X). Suppose you randomly selected 50 adults, and measured their mean arterial blood pressure (in mmHg) & body surface area (in m2). The results are reflected below: I = 1.6 s, = 0.5 y = 80.1 s, = 16.9 r 0.8 %3D %3D (Source: Nall, R. (2018, April 10). Mean arterial pressure: Normal, low, high readings plus treatment. Retrieved from https://www.healthline.com/health/mean-arterlal-pressurel There is a 99.7% chance that the actual MAP of those with body surface area of 1.66 m2 falls between and OA) 61.5; 102.0 B) 58.9; 104.5 O C) 66.5; 96.9 O D) 71.6; 91.9 E) 51.3; 112.1The manager of a fleet of automobiles is testing two brands of radial tires. She assigns one tire of each brand at random to the two rear wheels of eight cars and runs the cars until the tires wear out. The data are shown here (in kilometers). Find a 99% CI on the difference in mean life. Which brand would you prefer, based on this calculation? Car 1 2 3 4 Brand 1 36,925 45,300 36,240 32,100 Brand 2 34,318 42,280 35,500 31,950 Car Brand 1 5 37,210 6 48,360 7 8 Brand 2 38,015 47,800 38,200 37,810 33,500 33,215A photoconductor film is manufactured at a nominal thickness of 25 mils. The product engineer wishes to increase the mean speed of the film, and believes that this can be achieved by reducing the thickness of the film to 20 mils. Eight samples of each film thickness are manufactured in a pilot production process, and the film speed (in microjoules per square inch) is measured. For the 25-mil film, the sample data result is = 1.13 and 81 = 0.11, while for the 20-mil film, the data yield = 1.08 and 82 = 0.09. Note that an increase in film speed wwould lovwer the value of the observation in microjoules per square inch. (a) Do the data support the claim that reducing the film thickness increases the mean speed of the film? Use a = 0.10 and assume that the two population variances are equal and the underlying population of film speed is normally distributed. What is the P-value for this test? Round your answer to three decimal places (e.g. 98.765). The data v the claim that reducing the…
- A photoconductor film is manufactured at a nominal thickness of 25 mils. The product engineer wishes to increase the mean speed of the film, and believes that this can be achieved by reducing the thickness of the film to 20 mils. Eight samples of each film thickness are manufactured in a pilot production process, and the film speed (in microjoules per square inch) is measured. For the 25-mil film, the sample data result is = 1.15 and 81 = 0.11, while for the 20-mil film, the data yield 2 = 1.06 and 82 = 0.09. Note that an increase in film speed vould lower the value of the observation in microjoules per square inch. (a) Do the data support the claim that reducing the film thickness increases the mean speed of the film? Use a = 0.10 and assume that the two population variances are equal and the underlying population of film speed is normally distributed. What is the P-value for this test? Round your answer to three decimal places (e.g. 98.765). The data the claim that reducing the film…in uncam dIfferent from the mean amount of airborne bacteria in uncarpeted rooms. To test this, carpeted and ount of bacteria on the dish was recorded The results are presented in the following table: Uncarpeted rooms Carpeted rooms 12.1 9.8 Sample mean of bacteria per ft' Sample std. dev. of bacteria per ft Number of rooms tested 3.2 2.7 23 %3D U.05 to see test if the amount off airborne bacteria is different in carpeted rooms than in uncarpeted rooms. I. State your hypotheses II. State the allowable type I error rate (a.) III. Calculate your test statistic, the p-value and draw the physical representation of the p-value IV. Make your decision to either reject the null or fail to reject the null. V. Make a concluding statement. b.) Run a 95% confidence interval for the hypothesis test above. Don't forget to make a proper conclusion.