MGMT 650 Homework 11 pg 4

A data set includes 104 body temperatures of healthy adult humans having a mean of 98.7°F and a standard deviation of 0.66°F. Construct a 99% confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of 98.6°F as the mean body temperature? Click here to view at distribution table. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. What is the confidence interval estimate of the population mean µ? °F<μ<°F (Round to three decimal places as needed.) C… What does this suggest about the use of 98.6°F as the mean body temperature? O A. This suggests that the mean body temperature is lower than 98.6°F. O B. This suggests that the mean body temperature is higher than 98.6°F. O C. This suggests that the mean body temperature could very possibly be 98.6°F.

Consider the diagram below. 5.66 5.66 Which of the following represent the values of x, y, and z to the nearest hundredth? x = 9.78 y = 7,95

Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. Complete parts (a) through (c) below. TInterval (13.046,22.15) x = 17.598 Sx = 16.01712719 n = 50 Click the icon to view a t distribution table. a. What is the number of degrees of freedom that should be used for finding the critical value t/2? Critical t values df = (Type a whole number.) t Distribution: Critical t Values Left tail Area In One Tall 0.005 0.01 0.025 0.05 0.10 Degrees of Area In Two Talls Degrees of Freedom 0.01 0.02 0.05 0.10 0.20 Freedom 1 63.657 31.821 12.706 6.314 3.078 1 9.925 6.965 4.303 2.920 1.886 3 5.841 4.541 3.182 2.353 1.638 3 4 4.604 3.747 2.776 2.132 1.533 4 4.032 3.365 2.571 2.015 1.476 6 3.707 3.143 2.447 1.943 1.440 6 7 3.499 2.998 2.365 1.895 1.415 7 3.355 2.896 2.306 1.860 1.397 Critical t value 3.250 2.821 2.262 1.833 1.383 (negat ive) 10 3.169 2.764 2.228 1.812 1.372 10 11 3.106 2.718 2.201 1.796 1.363 11 12 3.055 2.681 2.179 1.782 1.356 12 13 3.012…

2.42 1.44 2.05 0.24 2.78 2.6 2.69 1.02 1.44 1.21 2.85 0.91 1.54 0.37 1.07 0.04 2.41 2.79 1.25 0.63 1.63 2.57 0.01 0.54 0.81 1.81 2.27 0.64 1.65 2.72 0.58 1.68 2.56 2.82 0.17 0.29 1.65 0.18 1.61 0.79 2.33 2.15 2.73 2.93 0.14 2.21 2.16 5. Find the mean of the data summarized in the given frequency distribution. Compare the computed mean to the actual mean of 52.3 miles per hour. Speed (miles per hour) Frequency 42-45 46- 49 50 - 53 54 -57 58-61 22 12 The mean of the frequency distribution is (Round to the nearest tenth as needed.) miles per hour. Which of the following best discribes the relationship between the computed mean and the actual mean? A. The computed mean is not close to the actual mean because the difference between the means is more than 5%. O B. The computed mean is close to the actual mean because the difference between the means is less than 5%. O C. The computed mean is close to the actual mean because the difference between the means is more than 5%. D. The computed…

Blue 0.143 Freckles 0.030 eyes No Freckles 0.857 0.182 Black 0.047 Green hair eyes 0.818 0.018 0.923 Brown eyes 0.982 0.278 0.061 Blue 0.259 eyes 0.722 0.232 Brown 0.327 Green hair eyes 0.768 0.153 Brown 0.461 0.414 eyes Caucasian 0.847 child 0.319 Blue 0.562 0.453 eyes 0.681 0.315 Green 0.302 Blonde hair eyes 0.698 0.164 Brown 0.123 eyes 0.025 0.836 0.857 Blue 0.473 eyes 0.143 0.767 Green Red hair 0.405 eyes 0.233 0.778 Brown 0.122 eyes 0.222

2.59 2.43 2.96 2.80 2.94 3.07 2.65 2.79 2.61 2.90 3.14 2.78 2.50 2.78 2.52 2.60 2.86 2.88 2.78 3.09 2.82 2.72 2.98 2.68 2.87 2.86 2.78 2.73 2.62 2.77 2.90 3.02 2.73 2.87 2.63 2.98 2.52 2.81 2.91 2.71 3.07 2.89 2.93 2.64 2.94 2.81 2.85 2.89 2.79 2.86 A school administrator believes that the mean class GPAs for a given course are higher than the preferred mean of 2.75 at a significance level of 0.05, and checks a randomly chosen sample of 50 classes. Create a histogram, and calculate x¯, the t-statistic, and the p-value. From the histogram, can normality be assumed? YES OR NO X= t= p= since the p-value  ( GREATER THAN OR LESS THAN)   the significance level 0.05, the null hypotheses is( fails to be rejected or  is rejected) (sufficient or insufficient evidence) exist that the mean GPA is greater than 2.75.

3-7. Write each answer with the correct number of significant figures (a) 1.02.1 + 3.4 + 5.8 12.300 0 (b) 106.9 31.4 (c) 107.868 (2.113 X 75.500 0 102(5.623 x 10 5 519.568 (d) (26.14/37.62) X 4.38 3.043 413 10 (e) (26.14/(37.62 x 10)) x (4.38 x 10 2) 3.043 413 x 10 11.933 7 (f) (26.14/3.38) + 4.2 (g) log(3.98 x 10*) = 4.897 79 X 10 4.599 9 (h) 10-6.31

A researcher interested in determining if there is an association between head circumference and brain volume tested for linear correlation at the a= 0.05 significance level. She took a random sample of twenty adults and measured the head circumference and brain volume (using ultrasound) for each person in the sample. The table above gives the paired data for the twenty adults. The researcher determined there is positive linear correlation between head circumference and brain volume at the 0.05 level of significance. With that information answer the following questions. 1.) Since we've determined there is linear correlation between head circumference and brain volume from the output on the TI83/84 what is "a" the y-intercept of the regression line? A. a=.508 B. a=34.740 C. a=-110.572 D. a=-823.829 2. Since we've determined there is linear correlation between head circumference and brain volume from the output on the T183/84 what is "b" the slope of the regression? A. b=34.740 B. b=.508…

Regular Coke Diet Coke Regular Pepsi Diet Pepsi 0.8192 0.7773 0.8258 0.7925 0.8150 0.7758 0.8156 0.7868 0.8163 0.7896 0.8211 0.7846 0.8211 0.7868 0.8170 0.7938 0.8181 0.7844 0.8216 0.7861 0.8247 0.7861 0.8302 0.7844 0.8062 0.7806 0.8192 0.7795 0.8128 0.7830 0.8192 0.7883 0.8172 0.7852 0.8271 0.7879 0.8110 0.7879 0.8251 0.7850 0.8251 0.7881 0.8227 0.7899 0.8264 0.7826 0.8256 0.7877 0.7901 0.7923 0.8139 0.7852 0.8244 0.7852 0.8260 0.7756 0.8073 0.7872 0.8227 0.7837 0.8079 0.7813 0.8388 0.7879 0.8044 0.7885 0.8260 0.7839 0.8170 0.7760 0.8317 0.7817 0.8161 0.7822 0.8247 0.7822 0.8194 0.7874 0.8200 0.7742 0.8189 0.7822 0.8172 0.7833 0.8194 0.7839 0.8227 0.7835 0.8176 0.7802 0.8244 0.7855 0.8284 0.7892 0.8244 0.7859 0.8165 0.7874 0.8319 0.7775 0.8143 0.7907 0.8247 0.7833 0.8229 0.7771 0.8214 0.7835 0.8150 0.7870 0.8291 0.7826 0.8152 0.7833 0.8227 0.7815 0.8244 0.7822 0.8211 0.7791 0.8207 0.7837…

A bank with branches located in a commercial district of a city and in a residential area has the business objective of developing an improved process for serving customers during the noon-to-1 P.M. lunch period. Management decides to first study the waiting time in the current process. The waiting time is defined as the number of minutes that elapses from when the customer enters the line until he or she reaches the teller window. Data are collected from a random sample of 15 customers at each branch. Complete parts (a) and (b) below. Click the icon to view the bank branch data. ... .. a. Assuming that the population variances from both banks are not equal, is there evidence of a difference in the mean waiting time between the two branches? (Use a = 0.1.) Let µ, be the mean waiting time of the commercial district branch and u, be the mean waiting time of the residential area branch. Determine the hypotheses. Choose the correct answer below. A. Ho: H1 = H2 H1: H1 # H2 В. Но: М 2 2 H1:…

Household   Paper Plastic1                    3.27     0.632                    2.41      1.133                   11.42    12.814                   7.98      6.095                   6.33.     3.866                   12.32    11.177                    6.96     7.608                  16.39     9.709                   6.83      3.5710                 5.86      3.9111                  9.41      3.3612                 6.44      8.4013                 14.33.   6.4314                 2.80      5.9215                 16.08    14.3616                 6.05       2.7317                 13.61      8.9518                 12.73.    14.8319                 9.55        9.2020                6.16         5.8821                 6.67        6.0922                 7.72        3.8623                15.09       9.1124                8.82        11.8925               11.08        12.4726               13.05        12.3127                9.83          6.2628                7.57           5.9229…

The 285 observations:   5.00 11.00 16.00 23.00 36.00 58.00 29.00 20.00 10.00  8.00  3.00  0.00  0.00  2.00 11.00 27.00 47.00 63.00 60.00 39.00 28.00 26.00 22.00 11.00 21.00 40.00 78.00122.00103.00 73.00 47.00 35.00 11.00  5.00 16.00 34.00 70.00 81.00111.00101.00 73.00 40.00 20.00 16.00  5.00 11.00 22.00 40.00 60.00 80.90 83.40 47.70 47.80 30.70 12.20  9.60 10.20 32.40 47.60 54.00 62.90 85.90 61.20 45.10 36.40 20.90 11.40 37.80 69.80106.10100.80 81.60 66.50 34.80 30.60  7.00 19.80 92.50154.40125.90 84.80 68.10 38.50 22.80 10.20 24.10 82.90132.00130.90118.10 89.90 66.60 60.00 46.90 41.00 21.30 16.00  6.40  4.10  6.80 14.50 34.00 45.00 43.10 47.50 42.20 28.10 10.10  8.10  2.50  0.00  1.40  5.00 12.20 13.90 35.40 45.80 41.10 30.10 23.90 15.60  6.60  4.00  1.80  8.50 16.60 36.30 49.60 64.20 67.00 70.90 47.80 27.50  8.50 13.20 56.90121.50138.30103.20 85.70 64.60 36.70 24.20 10.70 15.00 40.10 61.50 98.50124.70 96.30 66.60 64.50 54.10 39.00 20.60  6.70  4.30 22.70 54.80 93.80 95.80 77.20 59.10…

1.2 Round-off the following numbers to four significant figures: 38.46235, 0.70029, 0.0022218, 19.235101, 2.36425

= The accompanying table gives amounts of arsenic in samples of brown rice from three different states. The amounts are in micrograms of arsenic and all samples have the same serving size. The data are from the Food and Drug Administration. Use a 0.05 significance level to test the claim that the three samples are from populations with the same mean. Do the amounts of arsenic appear to be different in the different states? Given that the amounts of arsenic in the samples from Texas have the highest mean, can we conclude that brown rice from Texas poses the greatest health problem? Click the icon to view the data. L. 10 H1 H2 H3 H₁: At least one of the means is different from the others D. Ho: H1 H2 H3 H₁ H₁ H₂ H3 Determine the test statistic. PEDI The test statistic is (Round to two decimal places as needed.)

Use the following information to answer the next two exercises. The percent of licensed U.S. drivers (from a recent year) that are female is 48.60. Of the females, 5.03% are age 19 and under; 81.36% are age 20-64; 13.61% are age 65 or over. Of the licensed U.S. male drivers, 5.04% are age 19 and under; 81.43% are age 20-64; 13.53% are age 65 or over.

1. The total number of Corona cases in the US, please keep in mind it is a running total. Courtesy of worldometers.info Date in 2020 Feb. 15 Day ( Number of Corona Cases (C) 1 15 Feb. 26 12 68 Mar. 8 23 541 Mar. 19 34 14,896 171,655 517,100 Mar. 30 45 Apr10. Apr. 21 May 2 May 13 May 24 June 4 56 67 835,559 1,170,301 1,443,627 78 89 100 1,704,112 1,942,737 111 June 15 122 2,187,774 June 26 133 July 7 July 18 2,556,718 3,109,759 144 155 3,849,089 July 29 Aug. 9 Aug. 20 Aug. 31 166 4,588,252 177 5,228,738 5,771,205 188 199 6,239,373 Sept. 11 Sept. 22 210 6,673,451 221 7,109,730 a. Let C be the total number of Corona Virus cases in the United States by the end of the r days since Feruary 14, 2020. (Let r = 0 represent Feb. 15, 2020.) Create a scatter plot of the data. Take up the entire piece of graph paper. Make sure to label the axes, and title your graph. Does it look linear? Exponential, Quadratic? c. Use the calculator to find the regression equation for this data. Express the…