Answer:
Step-by-step explanation:
Null hypothesis:
[tex]H_o : \mu = 0.50[/tex]
Alternative hypothesis:
[tex]H_a : \mu > 0.50[/tex]
The sample proportion [tex]\hat p = \dfrac{x}{n}[/tex]
[tex]\hat p = \dfrac{46}{58}[/tex]
[tex]\hat p = 0.7931[/tex]
The test statistics:
[tex]Z = \dfrac{\hat p - p}{\sqrt{\dfrac{p(1-p)}{n} } }[/tex]
[tex]Z = \dfrac{0.7931 - 0.50}{\sqrt{\dfrac{0.5(1-0.5)}{58} } }[/tex]
[tex]Z = \dfrac{0.2931}{\sqrt{\dfrac{0.25}{58} } }[/tex]
[tex]Z = \dfrac{0.2931}{0.06565321643}[/tex]
Z = 4.46 (to two decimal places)
At 99% C.I
Level of significance is:
∝ = 1 - C.I
∝ = 1 - 0.99
∝ = 0.01
Thus, the lower confidence bound can be computed as:
[tex]= \hat p - Z_{\alpha} \sqrt{\dfrac{\hat p ( 1- \hat p )}{n} }[/tex]
Critical value of [tex]Z_{0.01/2} = 2.33[/tex]
[tex]= 0.7931 - 2.33 \sqrt{\dfrac{0.7931 (1-0.7931)}{58}}[/tex]
[tex]= 0.7931 - 2.33 (0.05319)[/tex]
= 0.7931 - 0.1239327
= 0.6691673
[tex]\simeq[/tex] 0.7 (to one decimal place)
(c)
Given:
[tex]P_o = 0.50[/tex] and [tex]\hat p = 0.80[/tex]
Then:
[tex]\beta (0.8) = \phi \Big [ \dfrac{p_o-\hat p + Z_{\alpha} \sqrt{\dfrac{p_o(1-p_o)}{n}}}{\sqrt{\dfrac{\hat p(1-\hat p)}{n} } } \Big][/tex]
[tex]\beta (0.8) = \phi \Big [ \dfrac{0.5-0.8 + 2.33 \sqrt{\dfrac{0.5(1-0.5)}{58}}}{\sqrt{\dfrac{0.8(1-0.8)}{58} }}\Big][/tex]
[tex]= \phi \Big [ \dfrac{-0.3+ 2.33 (0.06565)}{\sqrt{0.0027586 }} \Big][/tex]
[tex]= \phi \Big [ \dfrac{-0.3+ 0.1529645}{0.052522} \Big][/tex]
[tex]= \phi \Big [ \dfrac{-0.1470355}{0.052522 } \Big][/tex]
[tex]= \phi \Big [-2.799\Big][/tex]
= 0.0026 ( to four decimal place).