Question

A new fertilizer was applied to the soil of 146 bean plants. 23% showed increased growth. Find the margin of error and 95% confidence interval for the percentage of all bean plants which show increased growth after application of the fertilizer. Round all answers to 2 decimal places.

Answer 1

Answer:

The significance level for this case would be $\alpha=1-0.95=0.05$ and the critical value for this case would be:

$z_{\alpha/2}=1.96$

The margin of error is given by:

$ME = z_{\alpha/2} \sqrt{\frac{\hat p (1-\hat p)}{n}}$

And replacing we got:

$ME = 1.96 \sqrt{\frac{0.23*(1-0.23)}{146}} =0.0683$

And the margin of error for this case would be $ME = 0.07$

Step-by-step explanation:

For this case we have the following dataset given:

$n= 146$ represent the sample size

$\hat p =0.23$ represent the estimated proportion of interest

$Conf=0.95$ represent the confidence level

The significance level for this case would be $\alpha=1-0.95=0.05$ and the critical value for this case would be:

$z_{\alpha/2}=1.96$

The margin of error is given by:

$ME = z_{\alpha/2} \sqrt{\frac{\hat p (1-\hat p)}{n}}$

And replacing we got:

$ME = 1.96 \sqrt{\frac{0.23*(1-0.23)}{146}} =0.0683$

And the margin of error for this case would be $ME = 0.07$