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Sav [38]
3 years ago
5

In a study of the relationship of the shape of a tablet to its dissolution time, 6 disk-shaped ibuprofen tablets and 8 oval-shap

ed ibuprofen tablets were dissolved in water. The dissolve times, in seconds, were as follows:
Disk: 269.0, 249.3, 255.2, 252.7, 247.0, 261.6
Oval: 268.8, 260.0, 273.5, 253.9, 278.5, 289.4, 261.6, 280.2 Can you conclude that the mean dissolve times differ between the two shapes? Conduct a hypothesis test at the
α = 5% level.
a. State the appropriate null and alternative hypotheses.
b. Compute the test statistic.
c. Compute the P-value.
d. State the conclusion of the test in the context of this setting.
Mathematics
1 answer:
OLga [1]3 years ago
4 0

Answer:

Step-by-step explanation:

This is a test of 2 independent groups. Let μ1 be the mean dissolution time for disk-shaped ibuprofen tablets and μ2 be the mean dissolution time for oval-shaped ibuprofen tablets.

The random variable is μ1 - μ2 = difference in the mean dissolution time for disk-shaped ibuprofen tablets and the mean dissolution time for oval-shaped ibuprofen tablets.

We would set up the hypothesis.

a) The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

H1 : μ1 ≠ μ2 H1 : μ1 - μ2 ≠ 0

This is a two tailed test.

For disk shaped,

Mean, x1 = (269.0 + 249.3 + 255.2 + 252.7 + 247.0 + 261.6)/6 = 255.8

Standard deviation = √(summation(x - mean)²/n

n1 = 6

Summation(x - mean)² = (269 - 255.8)^2 + (249.3 - 255.8)^2 + (255.2 - 255.8)^2+ (252.7 - 255.8)^2 + (247 - 255.8)^2 + (261.6 - 255.8)^2 = 337.54

Standard deviation, s1 = √(337.54/6) = 7.5

For oval shaped,

Mean, x2 = (268.8 + 260 + 273.5 + 253.9 + 278.5 + 289.4 + 261.6 + 280.2)/8 = 270.7375

n2 = 8

Summation(x - mean)² = (268.8 - 270.7375)^2 + (260 - 270.7375)^2 + (273.5 - 270.7375)^2+ (253.9 - 270.7375)^2 + (278.5 - 270.7375)^2 + (289.4 - 270.7375)^2 + (261.6 - 270.7375)^2 + (280.2 - 270.7375)^2 = 991.75875

Standard deviation, s2 = √(991.75875/8) = 11.1

b) Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(x1 - x2)/√(s1²/n1 + s2²/n2)

Therefore,

t = (255.8 - 270.7375)/√(7.5²/6 + 11.1²/8)

t = - 3

c) The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [7.5²/6 + 11.1²/8]²/[(1/6 - 1)(7.5²/6)² + (1/8 - 1)(11.1²/8)²] = 613.86/51.46

df = 12

We would determine the probability value from the t test calculator. It becomes

p value = 0.011

d) Since alpha, 0.05 > than the p value, 0.011, then we would reject the null hypothesis. Therefore, we can conclude that at 5% significance level, the mean dissolve times differ between the two shapes

You might be interested in
2. The time between engine failures for a 2-1/2-ton truck used by the military is
OLEGan [10]

Answer:

A truck "<em>will be able to travel a total distance of over 5000 miles without an engine failure</em>" with a probability of 0.89435 or about 89.435%.

For a sample of 12 trucks, its average time-between-failures of 5000 miles or more is 0.9999925 or practically 1.

Step-by-step explanation:

We have here a <em>random variable</em> <em>normally distributed</em> (the time between engine failures). According to this, most values are around the mean of the distribution and less are far from it considering both extremes of the distribution.

The <em>normal distribution</em> is defined by two parameters: the population mean and the population standard deviation, and we have each of them:

\\ \mu = 6000 miles.

\\ \sigma = 800 miles.

To find the probabilities asked in the question, we need to follow the next concepts and steps:

  1. We will use the concept of the <em>standard normal distribution</em>, which has a mean = 0, and a standard deviation = 1. Why? With this distribution, we can easily find the probabilities of any normally distributed data, after obtaining the corresponding <em>z-score</em>.
  2. A z-score is a kind of <em>standardized value</em> which tells us the <em>distance of a raw score from the mean in standard deviation units</em>. The formula for it is: \\ z = \frac{x - \mu}{\sigma}. Where <em>x</em> is the value for the raw score (in this case x = 5000 miles).
  3. The values for probabilities for the standard normal distribution are tabulated in the <em>standard normal table</em> (available in Statistics books and on the Internet). We will use the <em>cumulative standard normal table</em> (see below).

With this information, we can solve the first part of the question.

The chance that a truck will be able to travel a total distance of over 5000 miles without an engine failure

We can "translate" the former mathematically as:

\\ P(x>5000) miles.

The z-score for x = 5000 miles is:

\\ z = \frac{5000 - 6000}{800}

\\ z = \frac{-1000}{800}

\\ z = -1.25

This value of z is negative, and it tells us that the raw score is 1.25 standard deviations <em>below</em> the population mean. Most standard normal tables are made using positive values for z. However, since the normal distribution is symmetrical, we can use the following formula to overcome this:

\\ P(z

So

\\ P(z

Consulting a standard normal table available on the Internet, we have

\\ P(z

Then

\\ P(z1.25)

\\ P(z1.25)

However, this value is for P(z<-1.25), and we need to find the probability P(z>-1.25) = P(x>5000) (Remember that we standardized x to z, but the probabilities are the same).

In this way, we have

\\ P(z>-1.25) = 1 - P(z

That is, the complement of P(z<-1.25) is P(z>-1.25) = P(x>5000). Thus:

\\ P(z>-1.25) = 1 - 0.10565

\\ P(z>-1.25) = 0.89435  

In words, a truck "<em>will be able to travel a total distance of over 5000 miles without an engine failure</em>" with a probability of 0.89435 or about 89.435%.

We can see the former probability in the graph below.  

The chance that a fleet of a dozen trucks will have an average time-between-failures of 5000 miles or more

We are asked here for a sample of <em>12 trucks</em>, and this is a problem of <em>the sampling distribution of the means</em>.

In this case, we have samples from a <em>normally distributed data</em>, then, the sample means are also normally distributed. Mathematically:

\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})

In words, the samples means are normally distributed with the same mean of the population mean \\ \mu, but with a standard deviation \\ \frac{\sigma}{\sqrt{n}}.

We have also a standardized variable that follows a standard normal distribution (mean = 0, standard deviation = 1), and we use it to find the probability in question. That is

\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

\\ z \sim N(0, 1)

Then

The "average time-between-failures of 5000" is \\ \overline{x} = 5000. In other words, this is the mean of the sample of the 12 trucks.

Thus

\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

\\ z = \frac{5000 - 6000}{\frac{800}{\sqrt{12}}}

\\ z = \frac{-1000}{\frac{800}{\sqrt{12}}}

\\ z = \frac{-1000}{230.940148}

\\ z = -4.330126

This value is so low for z, that it tells us that P(z>-4.33) is almost 1, in other words it is almost certain that for a sample of 12 trucks, its average time-between-failures of 5000 miles or more is almost 1.

\\ P(z

\\ P(z

\\ P(z

The complement of P(z<-4.33) is:

\\ P(z>-4.33) = 1 - P(z or practically 1.

In conclusion, for a sample of 12 trucks, its average time-between-failures of 5000 miles or more is 0.9999925 or practically 1.

7 0
2 years ago
PLEASE HELP ASAP!!
andriy [413]
If I did this correctly it simplifies to 0 = 8. So I think B is the correct answer. Hope this helps!!
6 0
3 years ago
The price of tickets for the dance was 1 ticket for $6 or 2 tickets for $8 she looked inside the cash box and found $200 and tic
Aleks04 [339]

We can say that exchanging one couple's ticket for an individual's ticket would increase the money in the cash box from 200 to 202 and it would result in an even number of couples tickets sold.

<u>Step-by-step explanation:</u>

Let the number of tickets sold to the individuals = s

Let the number of tickets sold to the couples = c

According to the question,

s + c = 46   (  Equation 1)

Since each individual's ticket is $6, the total amount of money made by selling tickets to individuals is 6s.

Similarly, since each ticket sold to couples is $8, the total amount of money made by selling tickets to couples is 8c.

So,

6 s + 8 c = 200        ( Equation 2)

On solving both the equations, we get

c = 38 and s = 8

Therefore, 8 tickets were sold to individuals and 38 tickets were sold to the couples.

5 0
2 years ago
Can anyone find the distance?
BigorU [14]
Sqrt((y2-y1)^2)+(x2-x1)^2)
6 0
2 years ago
What is 332,409 written in word form?
Diano4ka-milaya [45]

three hundred thirty-two thousand, four hundred nine.

:)

4 0
3 years ago
Read 2 more answers
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