Average Monthly Phone Bill and Sample Population Proportion - Statistics Project Example

Part A Step 1 Mean monthly bill=Total monthly bill/total users 9893/150=65.93 Step 2 Standard deviation=42.240 Step 3 Standard error=square root of total users/standard deviation √150/42.240=0.28 Step 4 Margin of error 0.28*2=0.579 Step 5 Confidence interval 0.579+65.93=66.50 0.579-65.93=65.35 Confidence interval between 65.35 and 66.50 We now have a 90% confidence interval of 65.35-66.50. Our best estimate of what the entire average monthly bill. Part B Confidence interval for population proportion @ 90% step 1 find sample proportion =no. of people in the sample/sample size 85/150=0.566 Step 2 Multiply p(1-p)/n 0.566(1-0.566)=0.2456 0.2456/150=0.00164 Step 3 find square root √0.00164=0.0404 Step 4 Multiply your answer by 1.645(margin of error) 0.0405*1.645=0.066 0.566+0.0666=0.632 0.599-0.0666=0.0994 Step 5 Confidence interval is between 0.0994 and 0.632 We now have a 90% confidence interval of 0.0994 - 0.632 Our best estimate of the proportion of users who are on postpaid plan. estimate Upper CL Lower CL Mean 61.5 66.43 65.47 proportion 65 63.2 50.02 Figure: CLUSTER COLUMN. Step 1 mean monthly bill =9893/150=65.93 Step 2 Standard deviation =42.240 Step 3 Standard error 42.240 √150/42.240=0.28 Step 4 Margin of error 0.28 *4.59=1.285222 step 5 5% significance 1.28522+65.93=67.21 1.28522-65.93=64.64 Step 1 5% significance is 4.59 Step 2 sample proportion=no. of people in sample /total sample size 85/150=0.566 Step 3 Multiply p(1-p)/total sample size 0.566(1-0.566)/150=0.00164 step 4 square root of 0.00164=0.0404 step 5 multiply 0.0404*4.59=0.1854 Step 6 plus or minus the margin of error 0.566+0.1854=0.7514 0.566-0.1854=0.3806 step 1 Male mean monthly bill 6654/111=59.94 Step 2 standard deviation 45 Step 3 standard error √111/45=0.234 Step 4 Margin of error 1.645*0,234=0.385 Step 5 Confidence interval 59.94+0.385=60.32 59.94-0.385=59.55 We now have a 90% confidence level @ 59.94-60.32. Our best estimate of entire male average monthly bill. Female monthly bill Step 1 mean =3239/39=83.05 Step 2 standard deviation 45 Step 3 standard error √39/45=0.1387 Step 4 Margin of error 1.645*0.1387=0.228 Step 5 Confidence interval 83.05-0.228=82.82 83.05+0.228=83.27 We now have a 90% confidence interval of 83.27-82.82 Our best estimate of entire female average monthly bill. conclusion female have a higher confidence interval than their male counter parts estimate upper CL lower CL male 59.94 60.32 59.55 female 83.05 83.27 82.82 figure 3 Difference of monthly bill for all male female users and all male users . Step 1 find Z a/2 by dividing the confidence interval by two, and looking that area up on the z-table 0.90/2=0.45 the closest Z score for 0.45 is 1.645 Step 2 Multiply step 1 by the standard deviation 1.645*45=74.025 Step 3 Divide step two by the margin of error 74.025/0.566=130.78 Step 4 Square step 3 (130.78*130.78=17105 Part B Step 1 Using the data given in the question, figure out the following variable Z a/2 ;divide the confidence interval by two, and look that area up in the z-table 0.90/2=0.45 The closest z-score for 0.45 is 1.645 E;(Margin of error):divide the given width by 2 4%/2=0.04/2 0.02 use the given percentage 56%=0.56 subtract from 1 1-0.56=0.44 Step 2 multiply p and set this no aside for a moment 0.56*0.44=0.2464 step 3 Divide Z a/2 by E 1.645/0.02=82.25 Step 4 square step 3 82.25*82.25=6756 Step 5 Multiply step 2 by step 4 0.56*0.44=0.2464 0.2464*6765=1666 1666 people to survey STATISTICAL DATA ANALYSIS Q1 (a). Confidence interval between 65.47-66.43 We now have a 90% confidence interval of 65.47-66.43 Our best estimate of the entire average monthly bill of all phone users. (b) Confidence interval between 0.5002-0.632 We now have a 90% confidence interval 0.5002-0.632.Our best estimate of the population proportion of all users who are on postpaid plan Q2 TTEST(One tailed test, two sample unequal variance) Critical t-value in your table using an alpha of 0.05 because you have a sampled=1-n,where n is the sample size of 150 the degree of freedom is 149(i.e. 1-150) T-table critical value table ,value of degree of freedom 149 and alpha of 0.05 T-statistics T-value=1.96 calculated p-value=0.19 Decision: critical t-statistics is higher than t critical table, reject null hypothesis that there is a significant difference that the average monthly bill is more than $61.5 of all users, therefore state the alternative hypothesis. interpreting: g the p-value=0.19 is higher than the critical value(greater than 0.05 significance) Conclusion: Accept the alternative hypothesis stating, there is no significant difference that the average monthly Bill is no more than $61.50 for all phone users. Q3 Decision: p Read More