I need help on number nine. Please answer ASAP this is homework needing to be turned in tommorow.

A study was conducted to determine whether an expectant mother's cigarette smoking has any effect on the bone mineral content of

svp [43]

Using Hypothesis testing,

a) Two samples are independent.

b)H₀: an expectant mother's cigarette smoking has any not effect on the bone mineral content of her otherwise healthy child.

Hₐ : an expectant mother's cigarette smoking has any effect on the bone mineral content of her otherwise healthy child.

c) Null hypothesis is accepted.

so, an expectant mother's cigarette smoking has no effect on the bone mineral.

We have given that,

A study which is conducted for check an expectant mother's cigarette smoking has any effect on the bone mineral content of her otherwise healthy child.

For new-born whose mother's cigarette smoking

sample size , n₁= 77

X-bar, x₁-bar = 0.098 g/cm

standard deviations, s₁ = 0.026 g/cm

For new-born whose mother did not cigarette smoking

sample size , n₂ = 161

standard deviations, s₂ = 0.025 g/cm.

mean (X-bar) , x₂-bar = 0.095 g/cm

a) the two samples are independent since they are different types of mothers, smoking mothers and non smoking mothers

b)H₀: an expectant mother's cigarette smoking has any not effect on the bone mineral content of her otherwise healthy child.

Hₐ: an expectant mother's cigarette smoking has any effect on the bone mineral content of her otherwise healthy child.

c) Test statistic:

t = (x₁-bar - x₂-bar )/S (√1/n₁+1/n₂)

where , S = √(s₁²(n₁ - 1) + s₂²(n₂ -1))/n₁+n₂ - 2

S = √((0.026)²(76) +(0.025 )²(160))/236

= 0.0253

then, t = (0.098 - 0.095 )/0.0253(√1/77 +1/161 )

=> t = 0.859

Using the critacal table critical t is

Critical t = ±1.970065

Degrees of freedom =236.0000

P-Value=0.3935 which is greater than α(0.05),

So , we accept H₀

Thus, an expectant mother's cigarette smoking has not effect on the bone mineral content of her otherwise healthy child.

To learn more about Hypothesis testing, refer:

brainly.com/question/4232174

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8 0

1 year ago

Kira's store collects 5% sales tax on every item sold. If she collected $22 in sales tax, what was the cost of the items her sto

Hatshy [7]

Answer:

48

Step-by-step explanation:

7 0

3 years ago

An apartment complex on Ferenginar with 250 units currently has 223 occupants. The current rent for a unit is 892 slips of Gold-

Triss [41]

Answer:

Step-by-step explanation:

Given data

Total units = 250

Current occupants = 223

Rent per unit = 892 slips of Gold-Pressed latinum

Current rent = 892 x 223 =198,916 slips of Gold-Pressed latinum

After increase in the rent, then the rent function becomes

Let us conside 'y' is increased in amount of rent

Then occupants left will be [223 - y]

Rent = [892 + 2y][223 - y] = R[y]

To maximize rent =

$\frac{dR}{dy}=0\\=2(223-y)-(892+2y)=0\\=446-2y-892-2y=0\\=-446-4y=0\\y=\frac{-446}{4}=-111.5$

Since 'y' comes in negative, the owner must decrease his rent to maximixe profit.

Since there are only 250 units available;

$y=-250+223=-27\\\\maximum \,profit =[892+2(-27)][223+27]\\=838 * 250\\=838\,for\,250\,units$

Optimal rent - 838 slips of Gold-Pressed latinum

8 0

3 years ago

What is volume of cylinder with 7 cm high and 3 cm radius

Alekssandra [29.7K]

Answer:

V≈461.81cm³

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A bee flies 20 feet per second directly to a flowerbed from its hive. The bee stays at the flowerbed for fifteen minutes. Then i

Ksenya-84 [330]

A. The equation to find the distance of the flowerbed from the hive is: $\frac{x}{20}+\frac{x}{12}+900= 1200$ , where x is the distance.

B. The flowerbed is 2250 feet far from the hive.

<u><em>Explanation</em></u>

Suppose, the distance of the flowerbed from the hive is $x$ feet.

The bee flies 20 feet per second directly to a flowerbed from its hive and flies directly back to the hive at 12 feet per second.

As, $Time=\frac{Distance}{Speed}$

So, the time taken by the bee to reach the flowerbed $= \frac{x}{20}$ seconds and the time taken to fly back to the hive $=\frac{x}{12}$ seconds.

The bee stays at the flowerbed for 15 minutes or (15×60)seconds or 900 seconds and it is away from the hive for a total of 20 minutes or (20×60)seconds or 1200 seconds.

So, the equation will be.......

$\frac{x}{20}+\frac{x}{12}+900= 1200\\ \\ \frac{3x+5x}{60}= 300\\ \\ 8x= 60*300\\ \\ 8x=18000\\ \\ x=\frac{18000}{8}=2250$

Thus, the distance of the flowerbed from the hive is 2250 feet.

5 0

3 years ago