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@ ieke a lest. Nnleckson-Google Chrome -Secu ht'ps www.mathsleom/Sudent, nlayeTes1.spx7leslu 163850813 icente,win yes During thc ter ct 2 U-2D29 th average 11 bll tor es den s ot a certan state as slus por month. A rancam sample ot 40 customers was sekcted during te w n er ot 2U y 201C and the avera e b les tour d to be S1/9. 2 ith a sample standard de lation ot2 /1 Complete pats a and b below. Submit Tost Ustamine tha hypor. t: 5pls possitle O D. Ho- 189 and H. . -189 The critical value(s) is(are iRound 10 tree dearna placcz as needed u52comma to separ te nawersas needed. I Hound to two dacimal placas as neadad) What conclusion should be dramn? O D. U not reject Ha 1 nerals not sufldent evdence to conclude tat tha avarage -tiity bil in th state was lower in the wintar ot zuus-2U10 than t was in tha winter ot 2ouu-zuus b) Does changing the value of tom 0.05 to 001 fiect your conclusion? "My or wny not? D. YRS, tha tandusion changes Ihara la not sntfciant tanca to rajat HD cUs cchnalogy to detemine the p lu frhis test. Hound to three declmal placs a neodad E O lype here to search

Explanation / Answer

xbar = 179.02
s = 20.73
n = 40

H0: mu >= 189
H1: mu < 189

alpha = 0.05
Critical value = -1.6849

Test statistics, t = (179.02 - 189)/(20.73/sqrt(40)) = -3.0448

As test statistics lie in rejection region, reject null hypothesis. Option C

If alpha changes from 0.05 to 0.01, critical value will be -2.4258
Still there are sufficient evidence to reject H0. Option A

p-value = 0.0020

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