Here are the survival times in days of 72 guinea pigs after they were injected with...

Here are the survival times in days of 72 guinea pigs after they were injected with infections bacteria in a medical experiment. Survival times, whether of machines under stress or cancer patients after treatment, usually have distributions that are skewed to the right.

43 45 53 56 56 57 58 66 67 73 74 79

80 80 81 81 81 82 83 83 84 88 89 91

91 92 92 97 99 99 100 100 101 102 102 102

103 104 107 108 109 113 114 118 121 123 126 128

137 138 139 144 145 147 156 162 174 178 179 184

191 198 211 214 243 249 329 380 403 511 522 598

a) Create a frequency distribution with 8 classes and describe its main features. Create a histogram from the data. Does the data show the expected right-skewed?

Expert Solution

orchestra answered 1 year ago

The survival times in days of 72 guinea pigs after they were injected with infectious bacteria...

The survival times in days of 72 guinea pigs after they were injected with infectious bacteria in a medical experiment is displayed in the table. 43 45 53 56 56 57 58 66 67 73 74 79 80 80 81 81 81 82 83 83 84 88 89 91 91 92 92 97 99 99 100 100 101 102 102 102 103 104 107 108 109 113 114 118 121 123 126 128 137 138 139 144 145 147 156...

Three groups of guinea pigs were injected respectively with 0.5mg, 1.0mg and 1.5mg of a tranquillizer,...

Three groups of guinea pigs were injected respectively with 0.5mg, 1.0mg and 1.5mg of a tranquillizer, and the following are the number of seconds it took them to fall asleep: 0.5mg 8.2 10 10.2 13.7 14 7.8 12.7 10.9 10.9 10.4 1.0mg 9.7 13.1 11 7.5 13.3 8.8 12.5 12.9 9.7 10.5 1.5mg 12 7.2 8 9.4 11.3 9 11.5 8.5 9.6 9.2 It is required to use the data to test whether differences in dosage have effect on the...

The data below are the survival times after treatment (in days) of some advanced colon cancer...

The data below are the survival times after treatment (in days) of some advanced colon cancer patients who were treated with ascorbate. 248, 377, 189, 1843, 180, 537, 519, 455, 406, 365, 942, 776, 372, 163, 101, 20, 283 (a) i) What is the point estimate for the average or mean of this data? ii) Report an appropriate 96% confidence interval estimate for the mean survival time after treatment of all advanced colon cancer patients who might be treated with...

The table below shows the means and standard deviations of the survival times (in days) of...

The table below shows the means and standard deviations of the survival times (in days) of some cancer patients who were treated with the same medication. Use these statistics to answer the question below. Type of Cancer Patient Mean Standard Deviation Breast 1395.9 1239.0 Bronchial 211.6 209.9 Colon 457.4 427.2 Ovarian 884.3 1098.6 Stomach 286.0 346.3 Which type of cancer patients (a) had the highest average survival time? (b) had the lowest average survival time? (c) had the most consistent...

The survival times in weeks for 20 male rats that were exposed to a high level...

The survival times in weeks for 20 male rats that were exposed to a high level of radiation are152 152 115 109 137 88 94 77 160 165 125 40 128 123 136 101 62 153 83 69. Data are from Lawless (1982) and are stored in the data frame RAT. Load the data with data(RAT)Construct a 95% confidence interval for the average survival time for male rats exposed to high levels of radiation. Round to two decimal places. Hint:...

Recorded here are the germination times (in days) for sixteen randomly chosen seeds of a new...

Recorded here are the germination times (in days) for sixteen randomly chosen seeds of a new type of bean: 11, 13, 12, 13, 20, 15, 18, 17, 14, 18, 10, 12, 13, 17, 14, 17 Send data to Excel Assuming that germination times are normally distributed, find a 95% confidence interval for the mean germination time for all beans of this type. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers...

Question 10 Recorded here are the germination times (in days) for fifteen randomly chosen seeds of...

Question 10 Recorded here are the germination times (in days) for fifteen randomly chosen seeds of a new type of bean: 11, 13, 11, 10, 15, 15, 10, 20, 10, 20, 13, 21, 16, 14, 12 Assuming that germination times are normally distributed, find a 95% confidence interval for the mean germination time for all beans of this type. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal...

In a study examining the survival of cancer patients, 100 patients were enrolled at baseline. After...

In a study examining the survival of cancer patients, 100 patients were enrolled at baseline. After one year, 20 patients were known to have died. Assuming no patients were lost to follow-up, what was the probability of surviving one year? (OK to show numerator/denominator)What would be the one-year survival rate in the previous question if ten patients had been lost to follow-up during the first year? Use the life table method.

Cholesterol levels were collected from patients two days after they had a heart attack (Ryan, Joiner...

Cholesterol levels were collected from patients two days after they had a heart attack (Ryan, Joiner & Ryan, Jr, 1985) and are in table #3.3.10. Find the five-number summary and interquartile range (IQR) (3points), and draw a box-and-whiskers plot with all the components labeled (3 points). Extra Credit (2points): State the shape of the distribution (1 point), upper and lower fence (2 points), and if there are any outliers (1 point). 16.An experiment is rolling a fair dieand then flipping...

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