Mathematics, 12.08.2020 20:01 kutie543

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% 1to1999%, would the margin of error for the confidence interval (mark one answer): decrease stay the same increase not enough information to answer
c. If the sample size in part a changed from 19 10 22. would the margin of errot for the confidence interval (mark one answer): decrease in stay the same increase in not enough information to answer
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.

Answers: 1

Show answers

Answers

Answer from: alexsrules9711

$336.04 < \mu < 443.96$

The margin of error will increase

The margin of error will decreases

The 99% confidence interval is 0.4107 < p < 0.4293

Step-by-step explanation:

From the question we are told that

The sample size n = 19

The sample mean is $\= x = \$\ 390$

The standard deviation is $\sigma = \$ \ 120$

Given that the confidence level is 95% then the level of significance is mathematically represented as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table

$Z_{\frac{\alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

=> $E = 1.96 * \frac{120}{\sqrt{19} }$

=> E = 53.96

The 95% confidence interval is

$\= x - E < \mu < \= x + E$

=> $390 - 53.96 < \mu < 390 - 53.96$

=> $336.04 < \mu < 443.96$

When the confidence level increases the $Z_{\frac{\alpha }{2} }$ also increases which increases the margin of error hence the confidence level becomes wider

Generally the sample size mathematically varies with margin of error as follows

$n \ \ \alpha \ \ \frac{1}{E^2 }$

So if the sample size increases the margin of error decrease

The sample proportion is mathematically represented as

$\r p = \frac{210}{500}$

$\r p = 0.42$

Given that the confidence level is 0.99 the level of significance is $\alpha = 0.01$

The critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{\alpha }{2} } = 2.58$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} }* \sqrt{ \frac{\r p (1- \r p )}{n} }$

=> $E = 0.42 * \sqrt{ \frac{0.42 (1- 0.42 )}{ 500} }$

=> E = 0.0093

The 99% confidence interval is

$\r p - E < p < \r p + E$

0.42 - 0.0093 < p < 0.42 + 0.0093

0.4107 < p < 0.4293

Answer from: Quest

8 apples are left.

step-by-step explanation:

10-2=8

10 apples

take 2

now u have 8

Answer from: Quest

b. 6

step-by-step explanation:

we are given the table representing an exponential function.

we can see from the table that, as x is increasing, the value of y is increasing in the power of 6.

i.e. the following relation is obtained,

x = 1, y = 6

x = 2, y = 36 = $6^{2}$

x = 3, y = 216 = $6^{3}$

x = 4, y = 1296 = $6^{4}$

x = 5, y = 7776 = $6^{5}$

so, the geometric series obtained is $6, 6^{2}, 6^{3}, 6^{4}, 6^{5}$ .

thus, the common ratio is given by $\frac{6^{2}}{6}$ = $\frac{36}{6}$ = 6.

hence, the common ratio is 6.

Another question on Mathematics

Mathematics, 21.06.2019 18:00

The swimming pool is open when the high temperature is higher than 20∘c. lainey tried to swim on monday and thursday (which was 3 days later). the pool was open on monday, but it was closed on thursday. the high temperature was 30∘c on monday, but decreased at a constant rate in the next 3 days. write an inequality to determine the rate of temperature decrease in degrees celsius per day, d, from monday to thursday.

Answers: 1

Answer

Mathematics, 21.06.2019 19:30

What is the lcm of 16 and 40 ? 24 and 40 ?

Answers: 2

Answer

Mathematics, 21.06.2019 20:30

Find the area of the triangle formed by the origin and the points of intersection of parabolas y=−3x^2+20 and y=x^2−16.

Answers: 3

Answer

Mathematics, 21.06.2019 21:50

Prism m and pyramid n have the same base area and the same height. cylinder p and prism q have the same height and the same base perimeter. cone z has the same base area as cylinder y, but its height is three times the height of cylinder y. the figures and have the same volume.

Answers: 3

Answer

You know the right answer?

The amount of money spent on textbooks per year for students is approximately normal. a. To estimat...

Questions

Mathematics, 03.02.2021 19:10

1. Classify 5x^5-7x+2x^2 degree, cubic b. quadratic quartic d. quintic...

Mathematics, 03.02.2021 19:10

They made $835 in ticket sales 145 people bought tickets child tickets were $4 adult tickets were $7 how many of each t...

Mathematics, 03.02.2021 19:10

AP CALCULUS What is the average value of y=3x^2+4x-5 on [0,2]? (don’t simplify please and show step by step work)...

Mathematics, 03.02.2021 19:10