What method do i use to solve this?

The tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm2 and a stand

Elanso [62]

Answer:

95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

Yes, this data suggest that the tensile strength was changed after the adjustment.

Step-by-step explanation:

We are given that the tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm 2 and a standard deviation of 2.15.

A machine was recently adjusted and a sample of 50 items were taken to determine if the mean tensile strength has changed. The mean of this sample is 74.28. Assume that the standard deviation did not change because of the adjustment to the machine.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean strength of 50 items = 74.28

$\sigma$ = population standard deviation = 2.15

n = sample of items = 50

$\mu$ = population mean tensile strength after machine was adjusted

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

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

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level of

significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $74.28-1.96 \times {\frac{2.15}{\sqrt{50} } }$ , $74.28+1.96 \times {\frac{2.15}{\sqrt{50} } }$ ]

= [73.68 kg/mm2 , 74.88 kg/mm2]

Therefore, 95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

Yes, this data suggest that the tensile strength was changed after the adjustment as earlier the mean tensile strength was 72 kg/mm2 and now the mean strength lies between 73.68 kg/mm2 and 74.88 kg/mm2 after adjustment.

8 0

3 years ago

please answer asap! A couple plans to have 5 children. The gender of each child is equally likely. Design a simulation involving

dimulka [17.4K]

Step-by-step explanation:

The gender of a child which is either a boy or a girl is determined by the XX-chromosomes, or XY-chromosomes.

Since the couple plan to have 5 children, the chance of a child being a boy is equal to the chance of it being a girl - the chances are 50/50.

What we do to achieve our aim is to run a simulation that would add an X or Y to an X for all 5 children.

Doing this 125 times, we obtain the number of trials we desire.

For each trial, we get for each child, C:

C1: X + (X or Y)

C2: X + (X or Y)

C3: X + (X or Y)

C4: X + (X or Y)

C5: X + (X or Y)

Since the chance of having an X is equal to the chance of having a Y, they equal probability, which is 0.5 for each.

6 0

2 years ago

In science class, the students observed 24 of the 300 frogs from the local aquarium. If 6 of the frogs observed had spots, how m

Alona [7]

Answer:

225 frogs

Step-by-step explanation:

Total population of frogs = 300 frogs.

Observed population of frogs = 24

6 of the 24 observed frogs had spots

Which means , the number of frogs that did not have spots = 24 - 6 = 18 frogs.

We were told to find how many of the total population can be predicted to NOT have spots. We would form a proportion.

If 24 frogs = 18 frogs with no spots

300 frogs = Y

Cross multiply

24Y = 300 × 18

Y = (300 × 18) ÷ 24

Y = 5400 ÷ 24

Y = 225 frogs.

This means out of 300 frogs, 225 frogs do not have spots.

Therefore, the total population that can be predicted to NOT have spots is 225 frogs.

the total population can be predicted to NOT have spots

3 0

2 years ago

The pentagons ABCDE and JKLMN are similar find the length x of JK

never [62]

The second shape is double the first so you would do AB = 2 and then times by 2 as its double and you will get 4
answer = 4

3 0

2 years ago

Read 2 more answers

Change 0.307 to a percent

Nookie1986 [14]

It would be 30.7% because when you change a decimal to a percent, the decimal point moves 2 places to the right. For a percent to a decimal, move the decimal point 2 places to the left

4 0

3 years ago