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1 year ago

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Is rap music more popular among young Indians than among young Asians? A sample survey compared 634 randomly chosen Indians aged

15 to 25 with 567 randomly selected Asians in the same age group. It found that 368 of the Indians and 130 Asians listened to rap music every day. Construct and interpret a 95% confidence interval for the difference between the proportions of black and white young people who listen to rap every day.

1 answer:

BabaBlast [244]1 year ago

8 0

Answer:

Step-by-step explanation:

Confidence interval for the difference in the two proportions is written as

Difference in sample proportions ± margin of error

Sample proportion, p = x/n

Where x = number of successes

n = number of samples

For the indians,

x = 368

n1 = 634

p1 = 368/634 = 0.58

For the asians

x = 130

n2 = 567

p2 = 130/567 = 0.23

Margin of error = z√[p1(1 - p1)/n1 + p2(1 - p2)/n2]

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.025 = 0.975

The z score corresponding to the area on the z table is 1.96. Thus, confidence level of 95% is 1.96

Margin of error = 1.96 × √[0.58(1 - 0.58)/634 + 0.23(1 - 0.23)/567]

= 1.96 × √0.00038422713 + 0.00031234568)

= 0.052

Confidence interval = 0.58 - 0.23 ± 0.052

Confidence interval = 0.35 ± 0.052

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Step-by-step explanation:

We are given that the average annual amount American households spend on daily transportation is $6312 (Money, August 2001). Assume that the amount spent is normally distributed.

(a) It is stated that 5% of American households spend less than $1000 for daily transportation.

Let X = <u><em>the amount spent on daily transportation</em></u>

The z-score probability distribution for the normal distribution is given by;

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$\frac{\$1000-\$6312}{\sigma}=-1.645$

$\sigma=\frac{-\$5312}{-1.645}$ = 3229.18

So, the standard deviation of the amount spent is $3229.18.

(b) The probability that a household spends between $4000 and $6000 is given by = P($4000 < X < $6000)

P($4000 < X < $6000) = P(X < $6000) - P(X $\leq$ $4000)

P(X < $6000) = P( $\frac{X-\mu}{\sigma}$ < $\frac{\$6000-\$6312}{\$3229.18}$ ) = P(Z < -0.09) = 1 - P(Z $\leq$ 0.09)

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P(X > x) = 0.03 {where x is the required range}

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