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

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The paper "Average Monthly Phone Bill and Sample Population Proportion" is an exceptional example of a statistical project on business. Some data relating to electronics and communication magazines requiring the performance of inferential techniques on the data set on previously analyzed statistical data on the mean monthly bill and sample population proportion, and the results found…
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Extract of sample "Average Monthly Phone Bill and Sample Population Proportion"

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
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(Average Monthly Phone Bill and Sample Population Proportion Statistics Project Example | Topics and Well Written Essays - 2250 words, n.d.)
Average Monthly Phone Bill and Sample Population Proportion Statistics Project Example | Topics and Well Written Essays - 2250 words. https://studentshare.org/business/2074579-business-analytics
(Average Monthly Phone Bill and Sample Population Proportion Statistics Project Example | Topics and Well Written Essays - 2250 Words)
Average Monthly Phone Bill and Sample Population Proportion Statistics Project Example | Topics and Well Written Essays - 2250 Words. https://studentshare.org/business/2074579-business-analytics.
“Average Monthly Phone Bill and Sample Population Proportion Statistics Project Example | Topics and Well Written Essays - 2250 Words”. https://studentshare.org/business/2074579-business-analytics.
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