the proportion of males who do not have children. To test her claim she randomly selects 346 females 58 did not have children. She then randomly selects 251 males workers and out of them 61 did not have children. Test her claim at œ=0.05 to see if she was right. The correct hypotheses are: O Ho:PF < PM HA:PF > PM(claim) Ο H0: pF > pM НА: PF <рм (claim) O Ho:PF НА: pF + рм (claim) = PM Since the level of significance is 0.05 the critical value is -1.645 The test statistic is: (round to 3 places) The p-value is: (round to 3 places) The decision can be made to: O reject Ho O do not reject Ho

College Algebra
10th Edition
ISBN:9781337282291
Author:Ron Larson
Publisher:Ron Larson
Chapter8: Sequences, Series,and Probability
Section8.7: Probability
Problem 4ECP: Show that the probability of drawing a club at random from a standard deck of 52 playing cards is...
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A student wishes to see if at her work place the proportion females who do not have children is less than
the proportion of males who do not have children. To test her claim she randomly selects 346 females 58
did not have children. She then randomly selects 251 males workers and out of them 61 did not have
children. Test her claim at a=0.05 to see if she was right. The correct hypotheses are:
О Но: pr < рм
НА: PF > рм(claim)
O Ho:PF 2 PM
На: рr <рм(claim)
О Но: pF — рм
HA:PF + PM(claim)
Since the level of significance is 0.05 the critical value is -1.645
The test statistic is:
(round to 3 places)
The p-value is:
(round to 3 places)
The decision can be made to:
O reject Ho
O do not reject Ho
Transcribed Image Text:A student wishes to see if at her work place the proportion females who do not have children is less than the proportion of males who do not have children. To test her claim she randomly selects 346 females 58 did not have children. She then randomly selects 251 males workers and out of them 61 did not have children. Test her claim at a=0.05 to see if she was right. The correct hypotheses are: О Но: pr < рм НА: PF > рм(claim) O Ho:PF 2 PM На: рr <рм(claim) О Но: pF — рм HA:PF + PM(claim) Since the level of significance is 0.05 the critical value is -1.645 The test statistic is: (round to 3 places) The p-value is: (round to 3 places) The decision can be made to: O reject Ho O do not reject Ho
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