I shall collect data from a population in order to estimate population parameters (e.g. and 2) by using estimating techniques

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I’m going to take the number of 13 to 16 year old students at Newmarket Upper School who took part in the questionnaire as my population as my first population and the number of 13 to 16 year olds in Suffolk as my second population.SAMPLE SIZE:To get my sample from the population, I will use a random number generator to get n responses from the question.

This would be the sample for my first population:How much pocket money do you normally get on weekly basis? (Tick appropriate box)Under�5Over�5To make accurate estimates of population parameters the sample must be large enough.The Central Limit Theorem allows predictions to be made about the distribution of the sample mean without any knowledge of the distribution of the parent population, as long as the sample is large enough.For this reason, my sample size, n will be set at 100, which I consider large enough for the distribution of its mean to be Normal. (According to the Central Limit Theorem).THE STATISTICAL THEORYCentral Limit Theorem:i) If the sample size is large enough, the distribution of the sample mean is approximately Normal.ii) The variance of the distribution of the sample mean is equal to the variance of the sample mean divided by the sample size; ?2n_Symbolically if X~ (unknown)(�, ?2) then X n ~N �, ? 2nThese approximations get closer as the sample size, n gets bigger.

(a good rule of thumb is n ? 30)HOW DATA WAS COLLECTED:374 students (male and female aged between 13 and 16 years old.) undertook a small-scale survey on matters such as the amount of pocket money they get on weekly basis, time spent on watching T.V. and part time jobs.For my question “How much pocket money do you normally get on weekly basis?” I recorded all the response I got as shown in the table below.MALEFEMALE13 yr oldsOVER �514 15 18 29 31 33 35 36 50 51 52 54 131 137 143 144 176 177 182200 204 2101 4 6 7 10 17 19 30 34 38 42 49 53 141 198 199 201 202 205 217 220223 224 225 229 231 234 239 246 249UNDER �513 16 32 37 41 55 119 135 136 145 179 180 206 207 235 242 243245 247 25039 40 42 43 48 138 178 213 214 226 227 230 232 236 237 248 37014 yr oldsOVER �58 20 56 69 71 83 86 94 99 105 109 113 120 140 146 152 155 158159 172 186 188 191 192 194 297 302 303 306 313 319 323 326329 335 338 340 342 348 349 350 351 353 357 360 3622 3 5 47 61 62 70 74 110 111 132 142 151 156 157 166 171 173 181 183185 187 196 208 209 211 228 241 293 296 298 299 300 301 309 314315 318 322 324 325 328 330 334 337 345 354 363 365 366 368UNDER �59 12 44 75 95 98 103 133 174 184 203 222 240 304 305 307 308310 317 329 332 333 341 343 347 355 366 37311 60 89 92 113 121 147 150 153 161 165 175 189 218 221 238 244294 311 312 316 327 331 336 352 356 358 361 364 365 371 372 37415 yr oldsOVER �558 65 68 72 73 76 79 81 83 90 117 124 126 134 149 163 164 170193 233 253 258 259 260 261 263 266 277 288 320 359 36759 67 78 80 82 84 85 91 96 106 107 112 114 118 123 125 128 129148 154 167 195 217 219 251 257 268 270 271 282 283 295 339UNDER �564 66 88 100 101 107 108 115 116 127 163 216 262 276 278 29032163 93 104 122 160 162 169 197 264 267 269 279 284 286 289 326346 367 36816 yr oldOVER �524 26 28 87 130 192 256 265 273 274 275 280 292252 272 281 291 369UNDER �521 22 23 25 57 77 97 215 254 255 285 28727 102 168 190 253Row data.

TOTALMALEFEMALE13 Y.O.14 Y.O.15 Y.

O16 Y.O.(374)(190)(184)(87)(152)(101)(34)OVER �563.30%32.60%31.

60%13.60%26.20%17.90%4.50%UNDER �537.

70%20.10%17.60%9.60%14.

40%9.10%4.50%WHAT CALCULATIONS WILL BE MADE USING THE DATA.1) The mean, standard deviation and variance of the sample.2) These will be used to estimate the variance and standard deviation of the parent population.

3) This will also be used to create confidence intervals for the mean of the parent population.4) Also, calculations that determine the size that a possible sample could be to achieve a certain percentage confidence interval for the mean to be a certain range.First I have to sort all the responses I got in numerical order so that I can use a random number generator to get my sample of 100 responses.13 YEAR OLDS14 YEAR OLDS15 YEAR OLDS16 YEAR OLDOVER �5UNDER �5OVER �5UNDER �5OVER �5UNDER �5OVER �5UNDER �5MFMFMFMFMFMFMFMF122324525372738990135136186187212213241242273274307308324325341342354355358359370371342526545574759192137138188189214215243244275276309310326327343344356357360361372373562728565776779394139140190191216217245246277278311312328329345346362363374782930585978799596141142192193218219247248279280313314330331347348364365910313260618081979814314419419522022124925028128231531633233334935036636711123334626382839910014514619619722222325125228328431731833433535135236836913143536646584851011021471481981992242252532542852863193203363373531516373866678687103104149150200201226227255256287288321322338339171839406869881051061511522022032282292572582892903233401920414270711071081531542042052302312592602912922122434410911015515620620723223326126229329445461111121571582082092342352632642952964748113114159160210211236237265266297298495011511616116223823926726829930051117118163164240269270301302119120165166271272303304121122167168305306123124169170125126171172127128173174129130175176131132177178133134179180181182183184185Using a random number generator, I picked 100 random responses as my sample.RESULTS (SAMPLE DATA):TOTALMALEFEMALE13 Y.

O.14 Y.O.15 Y.O16 Y.

O.(100)(48)(52)(26)(43)(23)(8)OVER �55424301223163UNDER �5462422142075To calculate the mean of the variance for the total number of 13-16 year old students who got pockect money on weekly basis:Let p be the proportion of student who get �5 or more.Although p is not yet know a good estimate will be: 54100Let X be the number of people in the sample who said they get more than �5 for their pocket money.Then X~(100,p)? to find mean for the sample (X) =np= 100xp=100pVariance = npq=100xpx(1-p)=100p(1-p)Provided that n, the size of sample is large enough (n ? 30), I can approximate this distribution with a Normal Distribution having the same mean and variance:X ~ N(�, ?2)X ~ N(100p, 100p(1-p))(but p = 0.

54)Mean = 100×0.54 � =54Variance =100×0.54×0.46 =24.

84 using the formula : n s2 ?2 = 25.09n – 1CONFIDENCE INTERVALS:To calculate how confident I am about the estimate of the population mean, I can use confidence intervals. These tell me how confident (as a percentage) I can be that the mean of the population falls within a given range.Using the formula :ps – 1.645 ps(1 – ps) ; p ; ps – 1.

645 ps (1 – ps)n nI will be able to find 90% confidence interval for the proportion of the population(p) who get �5 or more for their pocket money.0.54 – 1.645(V(0.

54×0.46)/100) ; p ; 0.54 + 1.645(V(0.54×0.

46)/100)0.54 – 0.0812 ; p ; 0.54 + 0.08120.

4588 ; p ; 0.6212between 45.9% and 62.1% (to 2 s.f.

)In other words, I am 90% cofidence that between 45% and 62% of the 374 students who answered the questionnaire get �5.00 or more for their pocket money.To find 95% the confidence interval:0.54 – 1.96(V(0.54×0.

46)/100) ; p ; 0.54 + 1.96(V(0.54×0.46)/100)0.

54 – 0.098 ; p ; 0.54 + 0.098between 44.2% and 63.

8% get �5.00 or more for their pocket money.FURTHER CONFIDENCE INTERVAL:If I wanted to be 99% confident;0.54 – 2.33(V(0.

54×0.46)/100) ; p ; 0.54 + 2.33(V(0.54×0.

46)/100)0.54 – 0.116 ; p ; 0.54 + 0.

116this would mean between 42.4% and 65.6% students who answered the questionnaire would get �5.00 or more for their pocket money.

PARAMETERS FOR THE SECOND POPULATION:To find the proprtion of 13 to 16 year old students in the Suffolk County who normally get �5.00 or more for their pocket money, I will use all the responses I got as my sample; rather than taking a sample from it.For this population random sampling is very hard to collect since this would involve questioning all 13 to 16 year olds in each individual school in the Suffolk County, and then randomly select my sample.For this reason I had to uses all of the responses as my sample.TOTALMALEFEMALE13 Y.

O.14 Y.O.15 Y.O16 Y.

O.(374)(190)(184)(87)(152)(101)(34)OVER �523011211851956717UNDER �5144786636573417To find the mean = np= 374 x (230/374)�= 230s2 = 374 x (0.615 x 0.385)= 88.6using n/(n – 1)s2 ; we have 374/(374-1) x 88.6?2 = 88.8Caluculating 90% confidence interval:0.615 – 1.645(V(0.615×0.385)/374) ; p ; 0.615 + 1.645(V(0.615×0.385)/374)0.574 ; p ; 0.656? between 57.4% and 65.6%To find the 95% confidence interval:0.615 – 1.96(V(0.615×0.385)/374) ; p ; 0.615 + 1.96(V(0.615×0.385)/374)0.566 ; p ; 0.664i.e between 56.6% and 66.4%CONCLUSION:Comparison of Cofidence IntervalsLooking at the two tables I can clearly see that the confidence intervals of the first popultion are bigger than those of the second population. This is because as I increased my sample size from 100 (first population) to 374 (second population), the value of the mean was getting closer to its true value.Going back to my original question “HOW MUCH POCKET MONEY DO YOU NORMARLLY GET ON WEEKLY BASIS?: OVER �5 OR UNDER �5” I can say that I am 95% confident that between 44% and 64% of all the student who participated in the qustionnaire get �5.00 or more every week for their pocket money.I am also 95% confident that between 57% and 66% of all 13 to 16 year olds in the Suffolk County receive �5 or more for their weekly pocket money.POSSIBLE EXTENSION:If I had enough time, I would also carry out the short survey to other schools within Suffolk.This would improve on my sampling of the second population and also would give me accurate population parameters; rather than using results from one school as the sample.LIMITATIONS:The size of the sample of the first population was small (compared to the popultion itself) ; therefore calculations that relied upon the data collected are therefore inaccurate to some extent.The results may be unreliable because some of the student might have given false information i.e lie about the amount of money they are given by their parents for pocket money.This could mean that the actual population parameters are somewhat different to the ones estimated here.

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