Answer:
The sample proportion p = 0.10
The claimed proportion p0 = 11%
Step-by-step explanation:
Step:-1
Given the claimed proportion [tex]P_{0}[/tex] = 11% = 0.11
[tex]Q_{0} = 1-P_{0} = 1-0.11 = 0.89[/tex]
The sample proportion of left handers is p = 0.10
The null hypothesis [tex]H_{0} = P_{0} = 0.11[/tex]
The alternative hypothesis [tex]H_{1} = P_{0} \neq 0.11[/tex]
Step 2:-
the test statistic
[tex]Z =\frac{p-P_{0} }{\sqrt{\frac{P_{0} Q_{0} }{n } } }[/tex]
substitute all values and simplification we get
[tex]Z =\frac{0.10-0.11 }{\sqrt{\frac{0.11(0.89) }{160 } } }[/tex]
After calculation we get
Z = -4.04
since modulus Z = 4.04 > 1.96 (tabulated value at 5% level)
we reject the null hypothesis is rejected at 5% level of significance
Conclusion:-
the claim proportion is not correct.
