i Excel File Edit View Insert Format Tools Data Window Help 58% Tue 8:10 PM a-E

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i Excel File Edit View Insert Format Tools Data Window Help 58% Tue 8:10 PM a-E Q Search Sheet Insert Page Layout Formulas Data Review View cut AutoSum, A Fill Calibri (Body) -11.A-Ar Wrap Text -Merge & Center" | $, % , % 4% conditional Format Cell Formatting as Table Sty es Delete Fermat Clear EEE Insert Delete Format Format Blu--.-'A, Filter E31 Note: Answer the Question(s) by Using the TI-3OXA (or other approved) Calculator IMPORTANT: Make sure you carry all decimals until you reach your answer, then round to four decimals. 12 14 15 16 17 18 19 In 2013, 45% of business owners gave a holiday gift to their employees. A 2014 survey of 43 business owners indicated that 31% plan to provide a holiday gift to their employees. Is there reason to believe the proportion of business owners providing gifts has decreased at -0.1? (See exercise 40 on page 418 of your textbook for a similar problem 21 23 For the hypothesis stated above. What is/are the critical value(s)? 25 26 Question 1 crit 1.28 What is the test statistic? 2.85 Question 3 What is the conclusion? 30 311 TINSETC 32 Click to Grade Your Work 36 37 38 39 40 Daily Problem+ Ready + 125% 17

Explanation / Answer

Given that,
possibile chances (x)=13.33
sample size(n)=43
success rate ( p )= x/n = 0.31
success probability,( po )=0.45
failure probability,( qo) = 0.55
null, Ho:p=0.45  
alternate, H1: p<0.45
level of significance, = 0.1
from standard normal table,left tailed z /2 =1.28
since our test is left-tailed
reject Ho, if zo < -1.28
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.31-0.45/(sqrt(0.2475)/43)
zo =-1.8453
| zo | =1.8453
critical value
the value of |z | at los 0.1% is 1.28
we got |zo| =1.845 & | z | =1.28
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: left tail - Ha : ( p < -1.84533 ) = 0.03249
hence value of p0.1 > 0.03249,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.45
alternate, H1: p<0.45
test statistic: -1.8453
critical value: -1.28
decision: reject Ho
p-value: 0.03249
we have enough evidence to support the claim

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