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3 years ago

PLEASE HELP! I CANT GET A BAD GRADE!

Mathematics

1 answer:

Alekssandra [29.7K]3 years ago

5 0

Answer:

It is decreased by 3.25

Step-by-step explanation:

When you add all 8 numbers you'll get 58 then you divide them by 8 you will get 7.25 which is the mean. To find out what is the decrease we need to take out the 30 and add all the other numbers and we get 28. After that you divide 28 by 7 and you get 4. Subtract 4 from 7.25 and you get 3.25.

2+2+2+3+5+6+8+30=58

58/8=7.25

2+2+2+3+5+6+8=28

28/7 = 4

7.25-4=3.25

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4 years ago

Read 2 more answers

James, a real estate appraiser is looking at homes in the 78681 zip code. He takes a random sample of 42 homes form the Austin M

vazorg [7]

Answer:

80% confidence interval for the proportion of homes that are "eclectic" is [0.59 , 0.78].

Step-by-step explanation:

We are given that James takes a random sample of 42 homes form the Austin MLS (that's where Realtors go for information) and finds that 29 of them are listed as "eclectic" style.

Firstly, the pivotal quantity for 80% confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of homes that are "eclectic" = $\frac{29}{42}$ = 0.69

n = sample of homes = 42

$\mu$ = population proportion of homes that are "eclectic"

Here for constructing 80% confidence interval we have used One-sample z proportion statistics.

So, 80% confidence interval for the population proportion, p is ;

P(-1.282 < N(0,1) < 1.282) = 0.80 {As the critical value of z at 10% level

of significance are -1.282 & 1.282}

P(-1.282 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.282) = 0.80

P( $-1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.80

P( $\hat p-1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.80

80% confidence interval for p = [ $\hat p-1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.69-1.282 \times {\sqrt{\frac{0.69(1-0.69)}{42} } }$ , $0.69+1.282 \times {\sqrt{\frac{0.69(1-0.69)}{42} } }$ ]