Question 1 A business researcher is testing the following hypotheses using a 10% level of significance.?H 0 : p = .70?H a : p A fail to reject the null hypothesis. B not enough information to conduct...

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Question 1

A business researcher is testing the following hypotheses using a 10% level of significance.?H0: p = .70?Ha: p



















Afail to reject the null hypothesis.
Bnot enough information to conduct the hypothesis test.
Ctake a larger sample.
Dreject the null hypothesis.



Question 2

A study by Hewitt Associates showed that 79% of companies offer employees flexible scheduling. Suppose a researcher believes that in accounting firms this figure is lower. The researcher randomly selects 415 accounting firms and through interviews determines that 303 of these firms have flexible scheduling. With a 1% level of significance, does the test show enough evidence to conclude that a significantly lower proportion of accounting firms offer employees flexible scheduling?
The value of the test statistic rounded to 2 decimal places is
z
=_______ and we a) reject the null hypothesis
b) fail to reject the null hypothesis
The tolerance is +/- 0.05.


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Quiz 1 Question 1 A business researcher is testing the following hypotheses using a 10% level of significance.?H0: p = .70?Ha: p <>






Quiz 1 Question 1 A business researcher is testing the following hypotheses using a 10% level of significance.
H0: p = .70
Ha: p < .70
a sample of 415 is taken and the sample proportion is .66. the business researcher's decision from this test is: a fail to reject the null hypothesis. b not enough information to conduct the hypothesis test. c take a larger sample. d reject the null hypothesis. question 2 a study by hewitt associates showed that 79% of companies offer employees flexible scheduling. suppose a researcher believes that in accounting firms this figure is lower. the researcher randomly selects 415 accounting firms and through interviews determines that 303 of these firms have flexible scheduling. with a 1% level of significance, does the test show enough evidence to conclude that a significantly lower proportion of accounting firms offer employees flexible scheduling? the value of the test statistic rounded to 2 decimal places is z =_______ and we a) reject the null hypothesis b) fail to reject the null hypothesis the tolerance is +/- 0.05. question 3 a business researcher wants to compare her observed distribution of frequency data to an expected distribution of data using the chi-square goodness-of-fit test. the data are given below. the degrees of freedom for this test are: observed expected 12 9 20 7 38 32 24 38 18 20 11 7 a 10 b 4 c 6 d 5 question 4 a business researcher wants to compare her observed distribution of frequency data to an expected distribution of data using the chi-square goodness-of-fit test. the data are given below. the observed chi-square value for this test is: observed expected 12 9 20 7 38 32 24 38 18 20 11 7 a 34.73 b 41.46 c 47.63 d 33.9 question 5 a researcher wants to test the following observed distribution of values to determine if the values are uniformly distributed. the researcher is using the chi-square goodness-of-fit test for this analysis. 213, 219, 209, 210, 216, 199, 217, 213 for α = .10, the researcher's decision is to: a reject the null hypothesis that the observed distribution is not uniform. b fail to reject the null hypothesis that the observed distribution is not uniform. c reject the null hypothesis that the observed distribution is uniform. d fail to reject the null hypothesis that the observed distribution is uniform. question 6 the following percentages come from a national survey of the ages of prerecorded-music shoppers. a local survey produced the observed values. does the evidence in the observed data indicate that we should reject the national survey distribution for local prerecorded-music shoppers? use   find the observed value of chi-square. round the answer to 2 decimal places. age precent from survey   10-14 9 22 15-19 23 50 20-24 22 43 25-29 14 29 30-34 10 19   35 22 49 the observed x (squared) = there is a) enough b) not enough evidence to declare that the distribution of observed frequencies is different from the distribution of expected frequencies. the tolerance is +/- 0.05. .70
a="" sample="" of="" 415="" is="" taken="" and="" the="" sample="" proportion="" is="" .66.="" the="" business="" researcher's="" decision="" from="" this="" test="" is:="" a="" fail="" to="" reject="" the="" null="" hypothesis.="" b="" not="" enough="" information="" to="" conduct="" the="" hypothesis="" test.="" c="" take="" a="" larger="" sample.="" d="" reject="" the="" null="" hypothesis.="" question="" 2="" a="" study="" by="" hewitt="" associates="" showed="" that="" 79%="" of="" companies="" offer="" employees="" flexible="" scheduling.="" suppose="" a="" researcher="" believes="" that="" in="" accounting="" firms="" this="" figure="" is="" lower.="" the="" researcher="" randomly="" selects="" 415="" accounting="" firms="" and="" through="" interviews="" determines="" that="" 303="" of="" these="" firms="" have="" flexible="" scheduling.="" with="" a="" 1%="" level="" of="" significance,="" does="" the="" test="" show="" enough="" evidence="" to="" conclude="" that="" a="" significantly="" lower="" proportion="" of="" accounting="" firms="" offer="" employees="" flexible="" scheduling?="" the="" value="" of="" the="" test="" statistic="" rounded="" to="" 2="" decimal="" places="" is="" z="_______" and="" we="" a)="" reject="" the="" null="" hypothesis="" b)="" fail="" to="" reject="" the="" null="" hypothesis="" the="" tolerance="" is="" +/-="" 0.05.="" question="" 3="" a="" business="" researcher="" wants="" to="" compare="" her="" observed="" distribution="" of="" frequency="" data="" to="" an="" expected="" distribution="" of="" data="" using="" the="" chi-square="" goodness-of-fit="" test.="" the="" data="" are="" given="" below.="" the="" degrees="" of="" freedom="" for="" this="" test="" are:="" observed="" expected="" 12="" 9="" 20="" 7="" 38="" 32="" 24="" 38="" 18="" 20="" 11="" 7="" a="" 10="" b="" 4="" c="" 6="" d="" 5="" question="" 4="" a="" business="" researcher="" wants="" to="" compare="" her="" observed="" distribution="" of="" frequency="" data="" to="" an="" expected="" distribution="" of="" data="" using="" the="" chi-square="" goodness-of-fit="" test.="" the="" data="" are="" given="" below.="" the="" observed="" chi-square="" value="" for="" this="" test="" is:="" observed="" expected="" 12="" 9="" 20="" 7="" 38="" 32="" 24="" 38="" 18="" 20="" 11="" 7="" a="" 34.73="" b="" 41.46="" c="" 47.63="" d="" 33.9="" question="" 5="" a="" researcher="" wants="" to="" test="" the="" following="" observed="" distribution="" of="" values="" to="" determine="" if="" the="" values="" are="" uniformly="" distributed.="" the="" researcher="" is="" using="" the="" chi-square="" goodness-of-fit="" test="" for="" this="" analysis.="" 213,="" 219,="" 209,="" 210,="" 216,="" 199,="" 217,="" 213="" for="" α=".10," the="" researcher's="" decision="" is="" to:="" a="" reject="" the="" null="" hypothesis="" that="" the="" observed="" distribution="" is="" not="" uniform.="" b="" fail="" to="" reject="" the="" null="" hypothesis="" that="" the="" observed="" distribution="" is="" not="" uniform.="" c="" reject="" the="" null="" hypothesis="" that="" the="" observed="" distribution="" is="" uniform.="" d="" fail="" to="" reject="" the="" null="" hypothesis="" that="" the="" observed="" distribution="" is="" uniform.="" question="" 6="" the="" following="" percentages="" come="" from="" a="" national="" survey="" of="" the="" ages="" of="" prerecorded-music="" shoppers.="" a="" local="" survey="" produced="" the="" observed="" values.="" does="" the="" evidence="" in="" the="" observed="" data="" indicate="" that="" we="" should="" reject="" the="" national="" survey="" distribution="" for="" local="" prerecorded-music="" shoppers?="" use=""  ="" find="" the="" observed="" value="" of="" chi-square.="" round="" the="" answer="" to="" 2="" decimal="" places.="" age="" precent="" from="" survey=""  ="" 10-14="" 9="" 22="" 15-19="" 23="" 50="" 20-24="" 22="" 43="" 25-29="" 14="" 29="" 30-34="" 10="" 19=""  ="" 35="" 22="" 49="" the="" observed="" x="" (squared)="There" is="" a)="" enough="" b)="" not="" enough="" evidence="" to="" declare="" that="" the="" distribution="" of="" observed="" frequencies="" is="" different="" from="" the="" distribution="" of="" expected="" frequencies.="" the="" tolerance="" is="" +/-="">
Answered Same DayDec 22, 2021

Answer To: Question 1 A business researcher is testing the following hypotheses using a 10% level of...

Robert answered on Dec 22 2021
124 Votes
Quiz 1
Question 1
A business researcher is testing the following hypotheses using a 10% level of significance.
H0: p = .70
Ha: p < .70
A sample of 415 is taken and the sample proportio
n is .66. The business researcher's decision from this test is:
    A
    fail to reject the null hypothesis.
    B
    not enough information to conduct the hypothesis test.
    C
    take a larger sample.
    D
    reject the null hypothesis.
Solution:
Null Hypothesis (H0): p = .70

Alternative Hypothesis (Ha): p < .70
Test Statistics
Z = (p’ – p)/ √ p (1 – p)/n
= (0.66 – 0.70)/ √0.70*0.30/415
= -1.78
Using Z-tables, the critical value is
Z (0.10) = -1.28
Since test statistics is less than the critical value, we reject the null hypothesis.
Question 2
A study by Hewitt Associates showed that 79% of companies offer employees flexible scheduling. Suppose a researcher believes that in accounting firms this figure is lower. The researcher randomly selects 415 accounting firms and through interviews determines that 303 of these firms have flexible scheduling. With a 1% level of significance, does the test show enough evidence to conclude that a significantly lower proportion of accounting firms offer employees flexible scheduling?
The value of the test statistic rounded to 2 decimal places is z =__-3.00_____ and we
a) reject the null hypothesis
b) fail to reject the null hypothesis
The tolerance is +/- 0.05.
Solution:
Null Hypothesis (Ho): p = 0.79
Alternative Hypothesis (Ha): p < 0.79
Sample proportion, p’ = X/n = 303/415 = 0.73
Test Statistics
Z = (p’ – p)/ √ p (1 – p)/n
= (0.73 – 0.79)/ √0.79*0.21/415
= -3.00
Using Z-tables, the critical value is
Z (0.01) = -2.33
Since test statistics is less than the critical value, we reject the null hypothesis.
Question 3
A business researcher wants to compare her...
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