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Radiology Physics Review. Computers (and statistics) David Paik david.paik@stanford.edu. Topics. Basic statistics Bits and bytes Images Computer systems. Basic Statistics. Null hypothesis: no difference exists p value: assuming null hypothesis , what is
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RadiologyPhysicsReview Computers (and statistics) David Paik david.paik@stanford.edu
Topics • Basic statistics • Bits and bytes • Images • Computer systems
Null hypothesis: no difference exists p value: assuming null hypothesis, what is the probability of this large a difference due purely to chance Common mistake: 99% probability of being true (Bzzt! WRONG!) Answer B • A paper states that a certain result (95% survival at 5 years) has a p value of 0.01. This means that: • A. The result is true 0.01% of the time. • B. There is a probability of 1 in 100 that this result was obtained by chance. • C. There is a 1% probability that the result is typical of the population. • D. None of the above.
Radioactive counts per unit time is Poisson distributed Poisson has mean = variance (mean=var=1000) Variance = (standard deviation)2 (st dev = 31.6) 68% is ±1 standard deviation 95% is ±2 standard deviation 1000 ± 32 is 968 to 1032 Answer B • A radioactive sample is counted many times, yielding a mean of 1000 counts. The most probable distribution is that 68% of the measurements fall between: • A. 990 and 1010 counts. • B. 968 and 1032 counts. • C. 936 and 1064 counts. • D. 900 and 1100 counts. • E. 800 and 1200 counts.
What does a positive predictive value (PPV) of 2% for abnormal screening mammography imply? • A. Radiologist calling too many abnormals. • B. Radiologist calling too many normals. • C. Image quality of images is poor. • D. Image quality of images is good. • E. Pathologists are doing a poor job. Sens = P(test+ | disease+) Spec = P(test- | disease-) PPV = P(disease+ | test+) NPV = P(disease- | test-) 2% of abnormal calls have disease 98% of abnormal calls have no disease Answer A
How many counts must be collected in a nuclear medicine well-counter with zero background to obtain an error limit of 1% with a confidence interval of 95%? • A. 1000. • B. 3162. • C. 10,000. • D. 40,000. • E. 100,000. Poisson: mean = variance 95% confidence interval = ±2 SD 2 x SD = 0.01 x mean 4 x variance = 0.0001 x mean2 4 x mean = 0.0001 x mean2 mean = 4/0.0001 = 40,000 Answer D
One byte of computer memory can store how many different possible integer values? • A. 2 • B. 16 • C. 256 • D. 512 • E. 65,536 Magic numbers to remember: 8 bits to a byte 28 = 256 2 byte words 216 = 65,536 Other useful ones: 210 = 1024 212 = 4096 Answer C
There are approximately ______ bits in a megabyte • A. 1024 • B. 2048 • C. 8000 • D. 2,000,000 • E. 8,400,000 1 megabyte = 1,000,000 bytes 8 bits per byte 8,000,000 bits per megabyte Don’t forget bit to byte conversion (sometimes, in binary people count kilo=1024 and mega=1,048,576) Answer E
If all 8 bits in a byte are set to one, then the decimal number is • A. 8 • B. 255 • C. 311 • D. 511 • E. 1023 Binary numbers start at 0 Thus, they go from 0 to 255 Remember the odometer analogy This is like an odometer at 999,999 miles Answer B
A terabyte of data corresponds to: • (a) 100 Gbytes • (b) 10,000 MBytes • (c) 1,000 Mbytes • (d) 1,000 GBytes • (e) 100,000 Mbytes mega, giga, tera Answer D
A non-subtracted angiographic image is acquired using a 1024x1024x16 bit matrix. The image is subsequently stored in PACS in a 512x512x16 bit format. Which of the following is true? ____ of the stored image is reduced in PACS • A. Only the contrast • B. Only the resolution • C. Only the apparent noise • D. Both the contrast and resolution • E. Both the resolution and apparent noise Resolution is reduced going from 1024x1024 to 512x512 Apparent noise is reduced if the pixels are averaged in a 2x2 neighborhood Answer E
A computed radiography image with 10-bit depth per pixel will have ___ shades of gray and ___ byte(s)/pixel. • A. 256, 1. • B. 256, 2. • C. 1024, 1. • D. 1024, 2. • E. 4049, 2. 210 = 1024 shades 8 bits per byte Thus 10 bits needs 2 bytes Extra 6 bits go unused Answer D
2 x 211 x 211 x 2 = 224 bytes 15 is almost 16 224/24 = 220 bytes after compression NOTE: speeds almost always bits/s not bytes/s 128 kbits/s = 16 kbytes/s = 214 bytes/s 220 bytes / 214 bytes/s = 26 s = 64 s Just over 1 minute Answer C • How much time would be required to transmit a pair of chest images (2048x2048 pixels, 2 bytes/pixel) to a radiologist’s home using ISDN (128 kbits/s) and a compression ratio of 15:1? • A. 20 minutes • B. 8 minutes • C. 1.2 minutes • D. 30 seconds • E. 10 seconds
A CT image consists of 200 slices, each 512 x 512 pixels, each pixel having 16-bit pixel depth. The size of the file is: • A. 500 kB • B. 5 MB • C. 10 MB • D. 50 MB • E. 100 MB Each CT slice: 512 x 512 x 2 = 524,288 bytes 0.5 megabyte 200 images x 0.5 megabyte/image = 100 megabytes Answer E
A 256 x 256 nuclear cardiology image can display 128 different colors. How many bytes of storage are needed? • A. 32 KB • B. 64 KB • C. 128 KB • D. 523 KB • E. 1 MB 128 = 27 Needs 1 byte/pixel 256 x 256 x 1 = 65,536 Answer B
How many megabytes (MB) of disk space are needed to store a single raw image file consisting of 512 x 512 pixels with each pixel having 16-bit pixel depth? • A. 0.25 MB • B. 0.5 MB • C. 0.75 MB • D. 1.00 MB • E. 1.25 MB 512 x 512 x 2 = 524,288 bytes 0.5 megabytes Answer B
How many bits are required to store 512 shades of gray? • A. 6 • B. 8 • C. 9 • D. 10 • E. 12 29 = 512 Answer C
How many 5122 images (16 bits per pixel) can be stored on a 2 GB disk? • A. 500 • B. 1000 • C. 4000 • D. 10000 • E. 50000 512 x 512 x 2 = 524,288 bytes 0.5 MB 2 GB = 2000 MB 4000 images Answer C
How much memory is needed to store a1 k2 radiograph with 256 shades of gray? • A. 0.1 MB • B. 1.0 MB • C. 10 MB • D. 100 MB • E. More than 100 MB 1000 x 1000 x 1 = 1,000,000 1 MB Answer B
Going from a 2562 image to a 5122 image will double the image’s: • A. spatial resolution • B. pixels • C. gray levels • D. transmission time • E. storage requirements Linear spatial resolution doubles in each dimension Pixels and size quadruple Gray levels stay same Answer A
The amount of time required to transfer 50 512 x 512 x 16 bit CT images over a 10 Mbit/s network is at best: • (a) 1 minute • (b) 20 seconds • (c) 200 seconds • (d) 2 seconds • (e) all night 512 x 512 x 2 = 0.5 MB 0.5 x 50 = 25 MB 10 Mb/s = 1.25 MB/s 25/1.25 = 20 s Answer B
CPU RAM Hard Disk Optical Disk Magnetic Tape Larger Faster • For the following questions, select one of the computer storage devices (answers may be used more than once) • A. RAM • B. Magnetic Tape • C. Hard Disk • D. CPU • E. Optical Disk • Which device has the longest access time? • Which device has the largest capacity? • Which storage device will lose information when the power goes off? • Which device can you generally not write over again?
The computational speed of a computer is measured in units of: • A. MB • B. MIPS • C. RVU • D. BAUD • E. BPI MIPS = millions of instructions per second FLOPS = floating point operations per second Answer B
ROM is memory that can be: • A. Used and changed freely • B. Freely read, but not written to • C. Repeatedly used to store output from an input device • D. Randomly accessed ROM = Read Only Memory RAM = Random Access Memory Answer B (but D too, bad question)
In Picture Archiving and Communication Systems (PACS), which of the following is a mass storage device? • A. DICOM • B. NEMA • C. RAID • D. ISDN • E. LAN RAID = Redundant Array of Independent Disks Striping makes access faster Mirroring adds redundancy Answer C
Points to Remember • P-value is probability of seeing that difference purely due to chance (not probability of being true) • ±1 st dev is 68% • ±2 st dev is 95% • Poisson statistics (counts per unit time) has mean=variance
Points to Remember • 1 byte = 8 bits • 8 bits stores 256 values • 2 bytes = 16 bits • 16 its stores 65,536 values • Transmission speeds usually in bits/sec not bytes/sec (don’t forget to convert between bits and bytes) • Convenient number: 210 = 1024 • Standard CT image (512x512, 2 bytes/pixel) = 0.5 MB