A block of copper weighs 2160g and has a volume of 240cm3. what is the density of the copper?

Suppose you are an expert on the fashion industry and wish to gather information to compare the amount earned per month by model

Ann [662]

Answer:

(1) The degrees of freedom for unequal variance test is (14, 11).

(2) The decision rule for the 0.01 significance level is;

If the value of our test statistics is less than the critical values of F at 0.01 level of significance, then we have insufficient evidence to reject our null hypothesis.

If the value of our test statistics is more than the critical values of F at 0.01 level of significance, then we have sufficient evidence to reject our null hypothesis.

(3) The value of the test statistic is 0.3796.

Step-by-step explanation:

We are given that you are an expert on the fashion industry and wish to gather information to compare the amount earned per month by models featuring Liz Claiborne's attire with those of Calvin Klein.

The following is the amount ($000) earned per month by a sample of 15 Claiborne models;

$3.5, $5.1, $5.2, $3.6, $5.0, $3.4, $5.3, $6.5, $4.8, $6.3, $5.8, $4.5, $6.3, $4.9, $4.2 .

The following is the amount ($000) earned by a sample of 12 Klein models;

$4.1, $2.5, $1.2, $3.5, $5.1, $2.3, $6.1, $1.2, $1.5, $1.3, $1.8, $2.1.

(1) As we know that for the unequal variance test, we use F-test. The degrees of freedom for the F-test is given by;

$\text{F}_(_n__1-1, n_2-1_)$

Here, $n_1$ = sample of 15 Claiborne models

$n_2$ = sample of 12 Klein models

So, the degrees of freedom = ( $n_1-1, n_2-1$ ) = (15 - 1, 12 - 1) = (14, 11)

(2) The decision rule for 0.01 significance level is given by;

If the value of our test statistics is less than the critical values of F at 0.01 level of significance, then we have insufficient evidence to reject our null hypothesis.

If the value of our test statistics is more than the critical values of F at 0.01 level of significance, then we have sufficient evidence to reject our null hypothesis.

(3) The test statistics that will be used here is F-test which is given by;

T.S. = $\frac{s_1^{2} }{s_2^{2} } \times \frac{\sigma_2^{2} }{\sigma_1^{2} }$ ~ $\text{F}_(_n__1-1, n_2-1_)$

where, $s_1^{2}$ = sample variance of the Claiborne models data = $\frac{\sum (X_i-\bar X)^{2} }{n_1-1}$ = 1.007

$s_2^{2}$ = sample variance of the Klein models data = $\frac{\sum (X_i-\bar X)^{2} }{n_2-1}$ = 2.653

So, the test statistics = $\frac{1.007}{2.653 } \times 1$ ~ $\text{F}_(_1_4,_1_1_)$

= 0.3796

Hence, the value of the test statistic is 0.3796.

3 0

2 years ago

Sunita bought a 10 kg box of grapes from the market. She gave 3 1/2 kg of her grapes to her friend Reena and 2 1/4 kg to Anita.

kirill115 [55]

Answer:

17/4 kg

Step-by-step explanation:

Sunita has = 10kg of grapes

Quantity given out :

To Reena =3 1/2

7/2 kg

To Anita =2 1/4

9/4 kg

Total quantity of grape given out

7/2+9/4

(14+9)/4

23/4

Quantity left after giving out = Total quantity of grape before giving out minus quantity of grape given out

10/1 - 23/4

40-23)/4

= 17/4 or 4 1/4

Hence, she has 17/4 kg of grape left with her

5 0

2 years ago

The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of 7 minutes and a stand

julsineya [31]

Answer:

There is a 2.28% probability that it takes less than one minute to find a parking space. Since this probability is smaller than 5%, you would be surprised to find a parking space so fast.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X.

Also, a probability is unusual if it is lesser than 5%. If it is unusual, it is surprising.

In this problem:

The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of 7 minutes and a standard deviation of 3 minutes, so $\mu = 7, \sigma = 3$ .

We need to find the probability that it takes less than one minute to find a parking space.

So we need to find the pvalue of Z when $X = 1$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1 - 7}{3}$

$Z = -2$

$Z = -2$ has a pvalue of 0.0228.

There is a 2.28% probability that it takes less than one minute to find a parking space. Since this probability is smaller than 5%, you would be surprised to find a parking space so fast.

5 0

2 years ago

The present age of Alex is ‘y’ years. Answer the following

skelet666 [1.2K]

Answer:

(a) After 5 years what will be his age?

"5 + y" years old

(b) What was his age 6 years back?

"y - 6" years old

(c) His grandfather‘s age is 5 times his age, What is the age of grandfather?

"5y" years old

(d) His father’s age is 6 years more than 3 times his age. What is his father’s age?