Question

13. In the planning stage, a sample proportion is estimated as
p^ = 36/60 = 0.60. Use this information to compute the minimum
sample size *n* required to estimate *p* with 95%
confidence if the desired margin of error *E* = 0.09. What
happens to *n* if you decide to estimate *p* with 90%
confidence?

**(You may find it useful to reference the z table. Round
intermediate calculations to at least 4 decimal places and
" z" value to 3 decimal places. Round up your answers to
the nearest whole number.)**

Confidence Level | n |

95% | ? |

90% | ? |

Please do not answer if you are not sure, THANK YOU! :)

Answer #1

13)

Solution :

Given that,

= 0.60

1 - = 0.40

margin of error = E = 0.09

**At 95% confidence**

Z_{/2}
= Z_{0.025} = 1.96

sample size = n = (Z_{
/ 2} / E )^{2} *
* (1 -
)

= (1.96 / 0.09)^{2} * 0.60 * 0.40

= 114

sample size = n = 114

**At 90% confidence**

Z_{/2}
= Z_{0.05} = 1.645

sample size = n = (Z_{
/ 2} / E )^{2} *
* (1 -
)

= (1.645 / 0.09)^{2} * 0.60 * 0.40

= 81

sample size = n = 81

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