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solniwko [45]
3 years ago
11

A random sample of 25 employees of a local company has been taken. A 95% confidence interval estimate for the mean systolic bloo

d pressure for all employees of the company is 123 to 139. Which of the following statements is valid?
(A) ​95% of the sample of employees has a systolic blood pressure between 123 and 139.
(B) ​If the sampling procedure were repeated many times, 95% of the resulting confidence intervals would contain the population mean systolic blood pressure.
(C) ​If the sampling procedure were repeated many times, 95% of the sample means would be between 123 and 139.
(D) ​95% of the population of employees has a systolic blood pressure between 123 and 139.
Mathematics
1 answer:
liraira [26]3 years ago
6 0

Answer:

(B) ​If the sampling procedure were repeated many times, 95% of the resulting confidence intervals would contain the population mean systolic blood pressure.

Step-by-step explanation:

A confidence interval of 95% means that there is a 95% certainty that for a given sample, the population mean will be within the confidence interval estimated.

This is the same as saying that if he sampling procedure were repeated many times, 95% of the time the population mean would be contained in the resulting confidence interval.

Therefore, the answer is B)

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Fewer hours

Step-by-step explanation:

2/3 is a fraction and shows the parts of a whole. This means that 2/3 of last week is fewer than last week. Last week is the whole for example 15 hours. This week will be 2/3 or 2/3*15 = 10 hours. This week will be less than last week.

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3 years ago
A sample of 114 mortgages approved during the current year showed that 36 were issued to a single-earner family or individual. T
kodGreya [7K]

Answer:

a-1) Reject H0 if zcalc > 1.645

a-2) z=\frac{0.316 -0.28}{\sqrt{\frac{0.28(1-0.28)}{114}}}=0.8561  

a-3) ii. False

b) ii. No

c) iii. n π > 10 and n(1 − π ) > 10

Step-by-step explanation:

1) Data given and notation

n=114 represent the random sample taken

X=36 represent the people that were issued to a single-earner family or individual

\hat p=\frac{36}{114}=0.316 estimated proportion of people that were issued to a single-earner family or individual

p_o=0.28 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

(a-1) H0: π ≤ .28 versus H1: π > .28. Choose the right option. Reject H0 if zcalc > 1.645 Reject H0 if zcalc < 1.645 a b

We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.28.:  

Null hypothesis:p\geq 0.28  

Alternative hypothesis:p < 0.28  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

The rejection zone would be on this case :

Reject H0 if zcalc > 1.645

Since is a right tailed test

(a-2) Calculate the test statistic. (Round intermediate calculations to 2 decimal places. Round your answer to 4 decimal places.) Test statistic

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.316 -0.28}{\sqrt{\frac{0.28(1-0.28)}{114}}}=0.8561  

(a-3) The null hypothesis should be rejected.

False, since our calculated value is less than our critical value we Fails to reject the null hypothesis

(b) Is this a close decision?

False the calculated value is significantly less than the critical value so we FAIL to reject the null hypothesis with enough confidence.

(c) State any assumptions that are required.

In order to satisfy the conditions we need the following two requirements:

iii. n π > 10 and n(1 − π ) > 10

And are satisfied:

114*0.28=31.92>10

114(1-0.28)=82,08>10

7 0
3 years ago
Please help me solve this equation 2x+y=10
krek1111 [17]
This is the answer
2x + y = 10 \\ y =  - 2x + 10

if substitute 1 and 2 and 3 as x then the points will be :
\binom{x}{y}   \:  \:  \:  \:  \:  \: \binom{1}{8} \:  \:  \:  \:  \:  \:  \binom{2}{6}   \:  \:  \:  \:  \:  \:  \binom{3}{4}


and that's the picture



good luck

7 0
3 years ago
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