Statistics- The 98% Ci for the difference between population

A1)a) The 98% Ci for the difference between population means is 🙁 x1 − x 2 ) ± Z 0.01 × S ×!1 1+n1 n2Where,x1 = 0.52, n1 = 10, s1 = 0.02x 2 = 0.5, n2 = 15, s 2 = 0.01Z 0.01 = 2.3263S=!2(n1 − 1) s12 + (n2 − 1) s 2n1 + n2 − 2Using all the values , the confidence interval is given by:(0.006, 0.034)b)Let ! µ1 denote the mean reaction time of non players of video games in the population and! µ 2 denote the mean reaction time of players of video games in the population.Null hypothesis, Ho: ! µ1 =! µ 2Alternative hypothesis, H1: ! µ1 >! µ 2( x1 − x 2 ) /{S ×The test statistic :Z= !11+ }n1 n2Where,x1 = 0.52, n1 = 10, s1 = 0.02x 2 = 0.5, n2 = 15, s 2 = 0.01S=!2(n1 − 1) s12 + (n2 − 1) s 2n1 + n2 − 2Thus Z=3.3226The tabulated value at 2.5 % LOS is 1.9596Thus HO is rejected as the calculated value is greater than tabulated value. Thus theresearcher can conclude his hypothesis.A2)a)The 97.5% CI for the difference in the population proportion is given by:ˆ ˆ( p1 − p 2 ) ± Z 0.0125 ×!ˆˆˆˆp1 × (1 − p1 ) p 2 × (1 − p 2 )+n1n2ˆˆWhere, ! p1 is the sample proportion of Americans and ! p 2 is the sample proportion ofCanadians.550= 0.551000225ˆp2 == 0.45500n1 = 1000ˆp1 =n2 = 500!Z 0.0125 = 2.2414Thus the CI is (.0389,0.1610)b)Let p1 and p2 be the population proportion for Americans and Canadiansrespectively/H0:p1=p2HI:p1! ≠ p2ˆ ˆ( p1 − p 2 ) /{Test stat: Z=!ˆˆˆˆp1 × (1 − p1 ) p 2 × (1 − p 2 )+}n1n2Using all the values from part a), Z =3.6698The calculated value is greater than tabulated value so we reject the null hypothesis.A3)Here we use the paired t test .Let xi’s denote the number of hours before program and yi’s denote the number ofhours after program. Then di’s denote the differences between xi’s and yi’s.dThe test statistic is t=! S d / n~ t n −1Where ,d =∑d!Sd =iin∑ (di− d )2in −1Thus using the given values , t=-1.04The tabulated value is 1.89Since abs t is less than tab t , we do no have sufficient evidence to reject thenull hypothesis. So we say there is no significant difference in the before and afterprogram mean hours.A4)Here we would set up a chi square testSEASONOBS FREQEXPECTED FREQ!Fall80701.4285Winter40502Spring70601.667Summer1020( o i − ei ) 2 / ei5The expected frequencies are calculated using the given percentage for each season andexpressing for 200 Like for Fall it is 35% and 35% of 200 is 70 and likewise for others. Now thetest statistic is 🙁 o i − ei ) 22χ =∑~ χ n −1eii!2So from the above table , the calculated value is 10.0955 (1.4285+2+1.667+5)Tabulated value is 7.814.Since calc value is greater than the tabulated one , we conclude thatobserved data contradicted the hypothesis.A5)We set up the contingency tables as below:OBSERVED VALUESOPINIONNEUTRAOPPOSED LUNDER 25INFAVOURTOTALOVER 555102010301050151053030AGE 35-5554525100!!EXPECTED VALUESOPINIONOPPOSEDUNDER 25AGENEUTRALIN FAVOURTOTAL622,512,55013,57,53030!!!209OVER 5551535-5594525100Expected values are calculated using{( row total *column total)/grand total}The testχ2 = ∑( o i − ei ) 2eiwith (r-1)*(c-1) df where is the # of rows and c is the #istatistic is given as: !of columns. Using the values from above and the formula, we get calculated value as17.352.The tabulated value at 1% LOS is: 13.2767.Since calculated value is greater than thetabulated value, we reject the hypothesis that there is no relation between age and opinionA6)a) The 95% CI for the population variance is given by:!(n − 1) S 2(n − 1) S 2>σ2 >χ (2 −α / 2 )χ (2α / 2 )1Here ,α = 0.05S 2 = 25n = 162χ 0.975,15 = 6.2622χ 0.025,15 = 27.4884!Using all the values, the CI is (13.64211,59.885)The CI for standard dev is(3.6935,7.7385)b)!HO : σ 2 = 1002! H 1 : σ < 100(n − 1) S 2~ χ (2n −1)σ2Test statistic:!Thus calculated value is: 3.75Tabulated value is 25.Since Calculated value is les than tabulated value we don’thave enough evidence to reject HOA7) Let !µ1 , µ 2 , µ 3 denote mean times to relieve pain for Brand A,B,C respectivelyH0: !µ1 = µ 2 = µ 3H1:Atleast 2 are differentWe run one way ANOVAAnova: Single FactorSUMMARYGroupsCountSumAverageVarianceBRAND A45112.75 4.916667BRAND B48120.2518.25BRAND C4761934ANOVASource ofVariationSSBetween Groups129.1667Within GroupsTotal171.5300.6667dfMSFP-valueF crit2 64.58333 3.389213 0.079947 5.7147059 19.0555611!From the table since the p value is greater than 0.025, we do not have sufficient evidence toreject the null hypothesis, thus the mean time is not statistically significant for 3 brands!

 

CLICK HERE TO ORDER A SIMILAR PAPER

We pride ourselves in writing quality essays

CLICK HERE TO CONTACT US

Lets Start Working

Plagiarism Free

We use anti-plagiarism software to ensure you get high-quality, unique papers. Besides, our writers have a zero plagiarism mentality

On Time Delivery

Your essay will be delivered strictly within the deadline.  If you have an urgent order, we can do it!

Money Back Guarantee

We offer warranty service, including free revisions, and a right to request a refund incase your expectations are not met!

THE BEST PAPER WRITER HELPER

Our Advantage

  • Say “NO” to plagiarism – FREE plagiarism report as an addition to your paper
  • The lowest prices that fit excellent quality
  • Authorship – you are the one who possesses the paper. We DO NOT re-sale or re-use any of them.

OUR PAPER WRITER HELPER GOODIES

Our Freebies

  • Free Cover Page
  • Free Revisions
  • Free Reference Page
  • Free 24/7 support

Pin It on Pinterest

Share This