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Dmitry_Shevchenko [17]
2 years ago
10

PLS HELP MARKING BRAINLIEST!

Mathematics
1 answer:
lukranit [14]2 years ago
3 0

Answer:

7 bagels and 4 donuts

Step-by-step explanation:

done it before

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4.One attorney claims that more than 25% of all the lawyers in Boston advertise for their business. A sample of 200 lawyers in B
AleksAgata [21]

Answer:

z=\frac{0.315 -0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}}=2.123  

p_v =P(Z>2.123)=0.0169  

The p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of lawyers had used some form of advertising for their business is significantly higher than 0.25 or 25% .  

Step-by-step explanation:

1) Data given and notation  

n=200 represent the random sample taken

X=63 represent the lawyers had used some form of advertising for their business

\hat p=\frac{63}{200}=0.315 estimated proportion of lawyers had used some form of advertising for their business

p_o=0.25 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)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that more than 25% of all the lawyers in Boston advertise for their business:  

Null hypothesis:p\leq 0.25  

Alternative hypothesis:p > 0.25  

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.

3) Calculate the statistic  

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

z=\frac{0.315 -0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}}=2.123  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

p_v =P(Z>2.123)=0.0169  

The p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of lawyers had used some form of advertising for their business is significantly higher than 0.25 or 25% .  

8 0
3 years ago
An SRS of 18 recent birth records at the local hospital was selected. In the sample, the average birth weight was 119.6 ounces a
Crazy boy [7]
The standard error of the mean is the standard deviation which is 6.50
8 0
3 years ago
A survey of 35 people was conducted to compare their self-reported height to their actual height. The difference between reporte
monitta

Answer:

The test statistic is t = 3.36.

Step-by-step explanation:

You're testing the claim that the mean difference is greater than 0.7.

At the null hypothesis, we test if it is 0.7 or less, that is:

H_0: \mu \leq 0.7

At the alternate hypothesis, we test if it is greater than 0.7, that is:

H_1: \mu > 0.7

The test statistic is:

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

0.7 is tested at the null hypothesis:

This means that \mu = 0.7

Survey of 35 people. From the sample, the mean difference was 0.95, with a standard deviation of 0.44.

This means that n = 35, X = 0.95, s = 0.44

Calculate the test statistic

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{0.95 - 0.7}{\frac{0.44}{\sqrt{35}}}

t = 3.36

The test statistic is t = 3.36.

7 0
3 years ago
Kini earns 9 dollars an hour at her job and her allowance is 8 dollars per week. Duke earns 7.50 dollars an hour and his allowan
Softa [21]

Answer:

Step-by-step explanation:

y = 9x + 8

y = 7.5x + 16

1.5x = 8

x = 5.3 hours

3 0
3 years ago
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Write the number for one million six hundred thousand .
shtirl [24]

Answer:

1,600,000

Step-by-step explanation:

4 0
3 years ago
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