P=,38 n= 500 Use the following to answer questions 10, 11, 12, 13, and 14. According to a report in USA Today, 38% of U.S. women wear size 7 shoes. A mail X=69 order catalog company took a random sample of 500 orders for women's shoes, and found that 169 of the orders were for size 7 shoes. 10. Set up the correct null and alternative hypotheses to test if these data indicate that the proportion of order size 7 shoes from the catalog is different from the proportion of women wearing size 7 shoes in the general population. H, : D=.38 vs H, :07.38 %3D 0=500 ×こ167 P =X=169 P= . 338 P= .38 G00 P=.34 9=.62 n 11. Calculate the value of the test statistic. Show your work and circle your answer. 2* - ア-P (. १५ - ,१8) 34-38 -1.84 %3D VC.35)(.62) 500 z* = - 1.84 a-025 12. Identify the rejection region, using a = 0.05 メ=.0a5 (a) z<-1.96 or z>1.96 (b) z>1.96 2- - 1,96 2-1.96 (c) z<-1.645 or z>1.645 2く -1.96 or 2>1.96 (d) z>1.645 (e) z<-1.28 or z>1.28 13. Find the p- value for this problem. Show your work and circle your answer. use fest Statistic P-value 0 329 0329 ,0329 P-value a 0658 •0658 2=-1,84 z*= 1.84 14. Comparing the p-value you found in question 13, with a=0.05, State your decision, draw your conclusion, and interpret your conclusion in the context of the problem. Decision: Ho Fail to'reject Conclusion: There is not sufticient evidence at the alpha level of Ô n O5 to conclude tthat the forportion_ot women g size 2 Shees id not 38go. wearing P-value = ,065s>a= 0.05 e value is qreater than alpha value so we fail to rejeet Ho

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P=•38 n= 500
Use the following to answer questions 10, 11, 12, 13, and 14.
According to a report in USA Today, 38% of U.S. women wear size 7 shoes. A mail X = 69
order catalog company took a random sample of 500 orders for women's shoes, and
found that 169 of the orders were for size 7 shoes.
10. Set up the correct null and alternative hypotheses to test if these data indicate that
the proportion of order size 7 shoes from the catalog is different from the
proportion of women wearing size 7 shoes in the general population.
Ho :
vs H. :
n=500
×こ167
P=
=.338 P=,38
500 Pニ,34262
%3D
=o62
11. Calculate the value of the test statistic. Show your work and circle your answer.
2* - ア-P
(. 3५ -.१8
%3D
-1.84
r638)(.62)
500
z=ー1.84
a-025
12. Identify the rejection region, using a =0.05
d=.0a5
(a) z<-1.96 or z>1.96
(b) z>1.96
2- - 1,96
2=1.96
(c) z<-1.645 or z>1.645
2く -1,96 orz71.96
(d) z>1.645
(e) z<-1.28 or z>1.28
13. Find the p – value for this problem. Show your work and circle your answer.
use test statistic
p-valne
• 0329
Value
t.0329
0658
•0658
.0329
0329
2=-1,84
z* = 1.84
(
P-value =
14. Comparing the p-value you found in question 13, with a=0.05, State your
decision, draw your conclusion, and interpret your conclusion in the context of
the problem.
Decision: Fail to 'reject Ho
Conclusion: There is not sufticient evidence at
the alpha level of Ooo5 to canclude that the
forfortion of women we
ig not 3890.
wearing
Size 2 shoes
P-value = ,045s>a= 0.05
e value is qeaten than
alpha value so we
fail to rejeet Ho
Transcribed Image Text:P=•38 n= 500 Use the following to answer questions 10, 11, 12, 13, and 14. According to a report in USA Today, 38% of U.S. women wear size 7 shoes. A mail X = 69 order catalog company took a random sample of 500 orders for women's shoes, and found that 169 of the orders were for size 7 shoes. 10. Set up the correct null and alternative hypotheses to test if these data indicate that the proportion of order size 7 shoes from the catalog is different from the proportion of women wearing size 7 shoes in the general population. Ho : vs H. : n=500 ×こ167 P= =.338 P=,38 500 Pニ,34262 %3D =o62 11. Calculate the value of the test statistic. Show your work and circle your answer. 2* - ア-P (. 3५ -.१8 %3D -1.84 r638)(.62) 500 z=ー1.84 a-025 12. Identify the rejection region, using a =0.05 d=.0a5 (a) z<-1.96 or z>1.96 (b) z>1.96 2- - 1,96 2=1.96 (c) z<-1.645 or z>1.645 2く -1,96 orz71.96 (d) z>1.645 (e) z<-1.28 or z>1.28 13. Find the p – value for this problem. Show your work and circle your answer. use test statistic p-valne • 0329 Value t.0329 0658 •0658 .0329 0329 2=-1,84 z* = 1.84 ( P-value = 14. Comparing the p-value you found in question 13, with a=0.05, State your decision, draw your conclusion, and interpret your conclusion in the context of the problem. Decision: Fail to 'reject Ho Conclusion: There is not sufticient evidence at the alpha level of Ooo5 to canclude that the forfortion of women we ig not 3890. wearing Size 2 shoes P-value = ,045s>a= 0.05 e value is qeaten than alpha value so we fail to rejeet Ho
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