Mathematics, 06.05.2020 02:39 jalst6084

An aquarium manager wants to study gift shop browsing. She randomly observes 120 couples that visit the aquarium with children and finds that 107 enter the gift shop at the end of their visit. She randomly observes 76 couples that visit the aquarium with no children and finds that 59 enter the gift shop at the end of their visit. Find the lower bound of the 95% confidence interval for the difference in population proportions of couples with children that enter the gift sho

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Answer from: 8848

95% confidence interval for the difference in population proportions of couples with children and with no children is [0.00134 , 0.219].

Step-by-step explanation:

We are given that an aquarium manager wants to study gift shop browsing.

She randomly observes 120 couples that visit the aquarium with children and finds that 107 enter the gift shop at the end of their visit. She randomly observes 76 couples that visit the aquarium with no children and finds that 59 enter the gift shop at the end of their visit.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportions is given by;

P.Q. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of couples that visit the aquarium with children who enters the gift shop at the end of their visit = $\frac{107}{120}$ = 0.89

$\hat p_2$ = sample proportion of couples that visit the aquarium with no children who enters the gift shop at the end of their visit = $\frac{59}{76}$ = 0.78

n_1 = sample of couples that visit the aquarium with children = 120

n_2 = sample of couples that visit the aquarium with no children = 76

Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.

So, 95% confidence interval for the difference between population proportions, is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ${(\hat p_1-\hat p_2)-(p_1-p_2)}$ < $1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

P( $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < (p_1-p_2)} < $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

95% confidence interval for (p_1-p_2)} =

[ $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ , $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ]

= [ $(0.89-0.78)-1.96 \times {\sqrt{\frac{0.89(1-0.89)}{120}+ \frac{0.78(1-0.78)}{76}} }$ , $(0.89-0.78)+1.96 \times {\sqrt{\frac{0.89(1-0.89)}{120}+ \frac{0.78(1-0.78)}{76}} }$ ]

= [0.00134 , 0.219]

Therefore, 95% confidence interval for the difference in population proportions of couples with children and with no children is [0.00134 , 0.219].

Answer from: Quest

146 days

step-by-step explanation:

1. (365*2)/5 =

730/5 =

146 days

2. 365/5=

73*2 = 146

Answer from: Quest

x+14

step-by-step explanation:

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An aquarium manager wants to study gift shop browsing. She randomly observes 120 couples that visit...