Question

several years ago, 40% of parents with children in grades k - 12 were satisfied with the quality of education the students receive. a recent poll found that 453 of 1,145 parents with children in grades k - 12 were satisfied with the quality of education the students receive. construct a 95% confidence interval to assess whether this represents evidence that parents attitudes toward the quality of education have changed.
click the icon to view the confidence interval critical value table.
what are the null and alternative hypotheses?
$h_0:p = 0.4$ versus $h_1:p
eq0.4$
(type integers or decimals. do not round.)
find the 95% confidence interval.
the lower bound is
the upper bound is
(round to three decimal places as needed.)

Explanation:

Step1: Calculate the sample proportion $\hat{p}$

The sample proportion $\hat{p}$ is given by $\hat{p}=\frac{x}{n}$, where $x = 453$ (number of successes) and $n=1145$ (sample size).
$\hat{p}=\frac{453}{1145}\approx0.396$

Step2: Find the critical value $z_{\alpha/2}$

For a 95% confidence interval, $\alpha = 1 - 0.95=0.05$, and $\alpha/2=0.025$.
From the standard normal distribution table, $z_{\alpha/2}=z_{0.025} = 1.96$

Step3: Calculate the margin of error $E$

The formula for the margin of error for a proportion is $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$
Substitute $\hat{p}\approx0.396$, $n = 1145$, and $z_{\alpha/2}=1.96$
$E=1.96\sqrt{\frac{0.396(1 - 0.396)}{1145}}$
First, calculate $0.396(1 - 0.396)=0.396\times0.604 = 0.239$
Then $\sqrt{\frac{0.239}{1145}}\approx\sqrt{0.000209}\approx0.0145$
$E=1.96\times0.0145\approx0.0284$

Step4: Calculate the lower and upper bounds

The lower bound is $\hat{p}-E$
$0.396- 0.0284=0.368$
The upper bound is $\hat{p}+E$
$0.396 + 0.0284=0.424$

Answer:

The lower bound is $0.368$
The upper bound is $0.424$