Answer :
Answer:
[tex]0.45 - 1.96\sqrt{\frac{0.45(1-0.45)}{36}}=0.2870.45 + 1.96\sqrt{\frac{0.45(1-0.45)}{36}}=0.613[/tex]
we can conclude at 95% of confidence that the true proportion of interest for this case is between 0.287 and 0.613
Step-by-step explanation:
The estimated proportion of interest is [tex]\hat p=0.45[/tex]
We need to find a critical value for the confidence interval using the normla standard distributon. For this case we have 95% of confidence, then the significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex].
And the critical value is:[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]
The confidence interval for the true population proportion is interest is given by this formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
Replacing the values provided we got:
[tex]0.45 - 1.96\sqrt{\frac{0.45(1-0.45)}{36}}=0.287\\\\0.45 + 1.96\sqrt{\frac{0.45(1-0.45)}{36}}=0.613[/tex]
And we can conclude at 95% of confidence that the true proportion of interest for this case is between 0.287 and 0.613
Answer:
0.45 +/- 1.645 x (sq rt 0.45 [1-0.45] ) / 36
Step-by-step explanation:
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