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4 years ago

Someone please help ASAP will give brainliest

Mathematics

2 answers:

mr_godi [17]4 years ago

7 0

Answer:

The answer is 15

Step-by-step explanation:

lorasvet [3.4K]4 years ago

3 0

Answer:15

Step-by-step explanation:

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zheka24 [161]

Solution :

$n_1=178, \hat p_1 = 0.48$

$n_2=427, \hat p_2 = 0.38$

$\hat p_1=\frac{x_1}{n_1}$

$x_1=n_1 \hat p_1$

= 178 x 0.48

= 85.44

≈ 85

$x_2=n_2 \hat p_2$

= 427 x 0.38

= 162.26

≈ 162

a). Yes, the sample sizes are large enough to use the large sample confidence interval so as to estimate the difference in the population proportions.

Let $\hat p_1 = 0.48, \ \ \hat p_2 = 0.38, n_1 = 178 \ \ n_2 = 427$

Where the $\text{subscript indicates}$ the age $18-29$ group and the 2 - subscript indicates the age $50-64$ group.

Since $n_1 \hat p_1 = 85.44$

$n_1(1- \hat p_1)=92.56, \ \ n_2\hat p_2 = 162.26, \ n_2(1-\hat p_2) = 264.74$

are all at least 10, the sample sizes are large enough to use the large sample confidence interval.

$P_0=\frac{x_1+x_2}{n_1+n_2}$

$=\frac{85.44+162.26}{178+427}$

= 0.409421

$P_0 = 0.4074$

$Q_0=1-P_0$

= 1 - 0.4094

= 0.5906

b). 90% confidence interval is

$=\left( \hat p_1- \hat p_2 \pm \frac{z \alpha}{2} \times \sqrt{P_0Q_0\left(\frac{1}{n_1}+\frac{1}{n_2}}\right) \right)$

$=\left( 0.48-0.38 \pm \frac{z \times 0.10}{2} \times \sqrt{0.4094 \times 0.5906\left(\frac{1}{178}+\frac{1}{427}}\right) \right)$

$=(0.1 \pm z 0.05 \times 0.04387082)$

$=(0.1 \pm 1.64 \times 0.0439)$

$=(0.1 - 0.071996, 0.1+0.071996)$

$=(0.028004, 0.171996)$

$=(0.0280, 0.1720)$

c). Zero is not included in the confidence interval. Answer is no. The difference in the two population proportion are different from each other.

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