Helpp ☹️☹️ plsplspsospsl

leonid [27]

Answer:

e)The information from the sample does not give enough information to support that more than 70% of Canadian children receive antibiotics during the first year of life

Step-by-step explanation:

The proportion we will use for the test is p = 70 %

From a random sample we got

n = 616

x = 438 then p₁ = 438/616 p₁ = 0,71 p₁ = 71 %

Then q₁ = 1 - p₁ q₁ = 1 - 0,71 q₁ = 0,29 q₁ = 29 %

a) Hypothesis test:

Null hypothesis H₀ p₁ = p

Alternative hypothesis Hₐ p₁ > p >70 %

CI = 95 % significance level α = 5 % α = 0,05

z(c) for α = 0,05 from z-table is z(c) = 1,64

b) To calculate z(s) = ( p₁ - p ) / √ p₁*q₁ / n

z(s) = ( 0,71 - 0,70 )/ √ 0,71*0,29/616

z(s) = 0,01 /0,01828

z(s) = 0,547 ≈ 0,55

c) p-value for z(s) is from z-table p-value ≈ 0,7088

d) p-value > 0,05 then we accept H₀ we don´t have enough evidence to reject H₀

e)The information from the sample does not give enough information to support that more than 70% of Canadian children receive antibiotics during the first year of life

3 0

3 years ago

camie and jorge each saving money for a new computer camie saves $5 per week plus $50 she received for her birthday. jorge saves

ki77a [65]

Answer:

7w > 5w + 50

Step-by-step explanation:

8 0

3 years ago

Find the area of the parallelogram shown below.

ale4655 [162]

Answer:

<h2>120 square units </h2>

Step-by-step explanation:

Given, 

Base of the parallelogram = 10

Height of the parallelogram = 12

Therefore, 

Area of the parallelogram = base × height 

= 10 × 12

= 120

Hence,

Area of the parallelogram is 120 square units (Ans)

5 0

3 years ago

Kendra had a beaded necklace business. She can make 12 necklaces in 2 hours. How long will it take her to make 9 necklaces?

mars1129 [50]

then + 10% of that total for overhead costs.

6 0

3 years ago

Read 2 more answers

Of 580580 samples of seafood purchased from various kinds of food stores in different regions of a country and genetically compa

Lubov Fominskaja [6]

Answer:

a) The 99% confidence interval would be given (0.204;0.296).

b) We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).

c) No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

Step-by-step explanation:

Part a

Data given and notation

n=580 represent the random sample taken

X represent the seafood sold in the country that is mislabeled or misidentified by the people

$\hat p=0.25$ estimated proportion of seafood sold in the country that is mislabeled or misidentified by the people

$\alpha=0.01$ represent the significance level (no given, but is assumed)

p= population proportion of seafood sold in the country that is mislabeled or misidentified by the people

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$0.25 - 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.204$

$0.25 + 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.296$

And the 99% confidence interval would be given (0.204;0.296).

Part b

We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).

Part c

A government spokesperson claimed that the sample size was too small, relative to the billions of pieces of seafood sold each year, to generalize. Is this criticism valid?

No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

4 0

4 years ago