Mathematics, 28.07.2020 17:01 hsjsjsjdjjd

A photoconductor film is manufactured at a nominal thickness of 25 mils. The product engineer wishes to increase the mean speed of the film and believes that this can be achieved by reducing the thickness of the film to 20 mils. Eight samples of each film thickness are manufactured in a pilot production process, and the film speed (in microjoules per square inch) is measured. For the 25-mil film, the sample data result is: Mean Standard deviation 1.15 0.11 For the 20-mil film the data yield: Mean Standard deviation 1.06 0.09 *Note: An increase in film speed would lower the value of the observation in microjoules per square inch. We may also assume the speeds of the film follow a normal distribution. Use this information to construct a 98% interval estimate for the difference in mean speed of the films. Does decreasing the thickness of the film increase the speed of the film?

Answers: 1

Show answers

Answers

Answer from: laxraAragon

A 98% confidence interval estimate for the difference in mean speed of the films is [-0.042, 0.222].

Step-by-step explanation:

We are given that Eight samples of each film thickness are manufactured in a pilot production process, and the film speed (in microjoules per square inch) is measured.

For the 25-mil film, the sample data result is: Mean Standard deviation 1.15 0.11 and For the 20-mil film the data yield: Mean Standard deviation 1.06 0.09.

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

P.Q. = $\frac{(\bar X_1 -\bar X_2)-(\mu_1- \mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ t__n_1_+_n_2_-_2

where, $\bar X_1$ = sample mean speed for the 25-mil film = 1.15

$\bar X_1$ = sample mean speed for the 20-mil film = 1.06

s_1 = sample standard deviation for the 25-mil film = 0.11

s_2 = sample standard deviation for the 20-mil film = 0.09

n_1 = sample of 25-mil film = 8

n_2 = sample of 20-mil film = 8

$\mu_1$ = population mean speed for the 25-mil film

$\mu_2$ = population mean speed for the 20-mil film

Also, $s_p =\sqrt{\frac{(n_1-1)s_1^{2}+ (n_2-1)s_2^{2}}{n_1+n_2-2} }$ = $\sqrt{\frac{(8-1)\times 0.11^{2}+ (8-1)\times 0.09^{2}}{8+8-2} }$ = 0.1005

Here for constructing a 98% confidence interval we have used a Two-sample t-test statistics because we don't know about population standard deviations.

So, 98% confidence interval for the difference in population means, ( $\mu_1-\mu_2$ ) is;

P(-2.624 < t_1_4 < 2.624) = 0.98 {As the critical value of t at 14 degrees of

freedom are -2.624 & 2.624 with P = 1%}

P(-2.624 < $\frac{(\bar X_1 -\bar X_2)-(\mu_1- \mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ < 2.624) = 0.98

P( $-2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ < $2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ < ) = 0.98

P( $(\bar X_1-\bar X_2)-2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ < ( $\mu_1-\mu_2$ ) < $(\bar X_1-\bar X_2)+2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ) = 0.98

98% confidence interval for ( $\mu_1-\mu_2$ ) = [ $(\bar X_1-\bar X_2)-2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ , $(\bar X_1-\bar X_2)+2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ]

= [ $(1.15-1.06)-2.624 \times {0.1005 \times \sqrt{\frac{1}{8}+\frac{1}{8} } }$ , $(1.15-1.06)+2.624 \times {0.1005 \times \sqrt{\frac{1}{8}+\frac{1}{8} } }$ ]

= [-0.042, 0.222]

Therefore, a 98% confidence interval estimate for the difference in mean speed of the films is [-0.042, 0.222].

Since the above interval contains 0; this means that decreasing the thickness of the film doesn't increase the speed of the film.

Answer from: Quest

$7.00 \cdot 10^{-7} m - 1\cdot 10^{-3} m$

explanation:

infrared radiation has wavelength in the range between 700 nm and 1 mm. lets' convert these values into scientific notation, keeping in mind that:

$1 nm = 10^{-9} m\\1 mm = 10^{-3} m$

let's start from the lower value, we have:

$700 nm = 700 \cdot 10^{-9} m = 7.00 \cdot 10^{-7} m$ (we moved the dot two places to the left, so we increased the exponent by 2 units, from -9 to -7)

the higher value is:

$1 mm = 1 \cdot 10^{-3} m$

so, the range of wavelength of infrared radiation is:

$7.00 \cdot 10^{-7} m - 1\cdot 10^{-3} m$

Answer from: Quest

Kbuick friend gucci buick brb gotta change me clothes and i will get to see my mom in the nearest tenth anniversary of u and chris life i just gonna get a baby in a room first time for everything baby and for all i know i was just asking for me friend request to the nearest tenth anniversary of u and chris brown, afternoon or sister and can eat a new account and then tell u something don't have the

Another question on Mathematics

Mathematics, 21.06.2019 13:30

Write the converse of the following statement: if the trees have no leaves, then it is fall. if the trees have no leaves, then it is fall. the trees have no leaves, therefore it is fall. it is fall since the trees have no leaves. if it is fall, then the trees have no leaves.

Answers: 2

Answer

Mathematics, 21.06.2019 17:00

How do businesses a country's economy? a. by investing in goods and services b. by increasing the unemployment rate c. by making profits

Answers: 2

Answer

Mathematics, 21.06.2019 21:30

Due to bad planning, two of the local schools have their annual plays on the same weekend. while 457 people attend one, and 372 attend the other, only 104 people were able to attend both. how many people went to at least one play?

Answers: 1

Answer

Mathematics, 21.06.2019 22:10

What is the factor form of x^2-9x+14

Answers: 2

Answer

You know the right answer?

A photoconductor film is manufactured at a nominal thickness of 25 mils. The product engineer wishes...

Questions