Mathematics, 12.08.2020 14:01 justloveyuki9667

The amount of money spent on textbooks per year for students is approximately normal. A. To estimate the population mean, 19 students are randomly selected the sample mean was $390 and the standard deviation was $120. Find a 95% confidence for the population meam.
B. If the confidence level in part a changed from 95% 1 to 1999%, would the margin of error for the confidence interval:
1. decrease.
2. stay the same.
3. increase not.
C. If the sample size in part a changed from 19% 10 to 22, would the margin of errot for the confidence interval:
1. decrease.
2. stay the same.
3. increase
D. To estimate the proportion of students who purchase their textbookslused, 500 students were sampled. 210 of these students purchased used textbooks. Find a 99% confidence interval for the proportion of students who purchase used text books.

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Answer from: pandadude123

(A) A 95% confidence for the population mean is [$332.16, $447.84] .

(B) If the confidence level in part (a) changed from 95% to 99%, then the margin of error for the confidence interval would increase.

(C) If the sample size in part (a) changed from 19 to 22, then the margin of error for the confidence interval would decrease.

(D) A 99% confidence interval for the proportion of students who purchase used textbooks is [0.363, 0.477] .

Step-by-step explanation:

We are given that 19 students are randomly selected the sample mean was $390 and the standard deviation was $120.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ t_n_-_1

where, $\bar X$ = sample mean = $390

s = sample standard deviation = $120

n = sample of students = 19

$\mu$ = population mean

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.101 < t_1_8 < 2.101) = 0.95 {As the critical value of t at 18 degrees of

freedom are -2.101 & 2.101 with P = 2.5%}

P(-2.101 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.101) = 0.95

P( $-2.101 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.101 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.101 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.101 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.101 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.101 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $\$390-2.101 \times {\frac{\$120}{\sqrt{19} } }$ , $\$390+2.101 \times {\frac{\$120}{\sqrt{19} } }$ ]

= [$332.16, $447.84]

(A) Therefore, a 95% confidence for the population mean is [$332.16, $447.84] .

(B) If the confidence level in part (a) changed from 95% to 99%, then the margin of error for the confidence interval which is $Z_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }$ would increase because of an increase in the z value.

(C) If the sample size in part (a) changed from 19 to 22, then the margin of error for the confidence interval which is $Z_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }$ would decrease because as denominator increases; the whole fraction decreases.

(D) We are given that to estimate the proportion of students who purchase their textbooks used, 500 students were sampled. 210 of these students purchased used textbooks.

Firstly, the pivotal quantity for finding the 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 students who purchase their used textbooks = $\frac{210}{500}$ = 0.42

n = sample of students = 500

p = population proportion

Here for constructing a 99% confidence interval we have used a One-sample z-test statistics for proportions

So, 99% confidence interval for the population proportion, p is ;

P(-2.58 < N(0,1) < 2.58) = 0.99 {As the critical value of z at 0.5%

level of significance are -2.58 & 2.58}

P(-2.58 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.58) = 0.99

P( $-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

99% confidence interval for p = [ $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.42 -2.58 \times {\sqrt{\frac{0.42(1-0.42)}{500} } }$ , $0.42 +2.58 \times {\sqrt{\frac{0.42(1-0.42)}{500} } }$ ]