Question

Several years​ ago, 39​% of parents who had children in grades​ K-12 were satisfied with the quality of education the students receive. A recent poll asked 1 055 parents who have children in grades​ K-12 if they were satisfied with the quality of education the students receive. Of the 1 055 ​surveyed, 466 indicated that they were satisfied. Construct a 95​% confidence interval to assess whether this represents evidence that​ parents' attitudes toward the quality of education have changed.

Answer 1

Answer with explanation:

Let p be the population proportion  of parents who had children in grades​ K-12 were satisfied with the quality of education the students receive.

Given : Several years​ ago, 39​% of parents who had children in grades​ K-12 were satisfied with the quality of education the students receive.

Set hypothesis to test :

$H_0: p=0.39\\\\H_a :p\neq0.39$

Sample size : n= 1055

Sample proportion : $\hat{p}=\dfrac{466}{1055}=0.441706161137\approx0.44$

Critical value for 95% confidence : $z_{\alpha/2}=1.96$

Confidence interval : $\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}$

$0.44\pm (1.96)\sqrt{\dfrac{0.39(1-0.39)}{1055}}\\\\0.44\pm0.0299536805135\\\\0.44\pm0.03\\\\=(0.44-0.03, 0.44+0.03)\\\\=(0.41,\ 0.47)$

Since , Confidence interval does not contain 0.39.

It means we reject the null hypothesis.

We conclude that 95​% confidence interval represents evidence that​ parents' attitudes toward the quality of education have changed.

Answer 2

We can say that 95% confidence interval represents evidence that​ parents attitudes toward the quality of education have changed.

Step-by-step explanation:

Given :

39​% of parents who had children in grades​ K-12 were satisfied with the quality of education the students receive.

The 1055 ​surveyed, 466 indicated that they were satisfied with the quality of education the students receive.

95% confidence interval.

Sample Size, n = 1055

Solution :

Hypothesis -

$\rm H_0 : p = 0.39$

$\rm H_a : p \neq 0.39$

Now sample proportion,

$\hat{p} = \dfrac{466}{1055}$

$\hat{p} \approx 0.44$

For 95% confidence critical value is

$z_\frac{\alpha}{2} = 1.96$

Now Confidence interval is

$\hat{p} \pm z_\frac{\alpha }{2}\sqrt{\dfrac{p(1-p)}{n}} = 0.44 \pm (1.96)\sqrt{\dfrac{0.39(1-0.39)}{1055}}$

$= 0.44 \pm (1.96)\sqrt{\dfrac{0.39\times(0.61)}{1055}} = 0.44 \pm 0.03$

$\rm = 0.44-0.03 \; \; ,\; \; 0.44+0.03$

= (0.41 , 0.47)

Since confidence interval does not contain 0.39 hence we reject the null hypothesis.

Therefore we can say that 95% confidence interval represents evidence that​ parents attitudes toward the quality of education have changed.

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https://brainly.com/question/12905909?referrer=searchResults