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Julli [10]
1 year ago
8

From Statistics and Data Analysis from Elementary to Intermediate by Tamhane and Dunlop, pg 265. A thermostat used in an electri

cal device is to be checked for accuracy of its design setting of 200◦F. Ten thermostats were tested to determine their actual settings, resulting in the following data: 202.2 203.4 200.5 202.5 206.3 198.0 203.7 200.8 201.3 199.0 Perform the t-test to determine if the mean setting is different from 200◦F. Use α = 0.05
Mathematics
1 answer:
Leviafan [203]1 year ago
4 0

Answer:

t=\frac{201.77-200}{\frac{2.41}{\sqrt{10}}}=2.32    

The degrees of freedom are given by:

df=n-1=10-1=9  

The p value for this case is given by:

p_v =2*P(t_{(9)}>2.32)=0.0455    

For this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 200 F.

Step-by-step explanation:

Information given

data: 202.2 203.4 200.5 202.5 206.3 198.0 203.7 200.8 201.3 199.0

We can calculate the sample mean and deviation with the following formulas:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

\sigma=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X=201.77 represent the sample mean    

s=2.41 represent the sample standard deviation    

n=10 sample size    

\mu_o =200 represent the value that we want to test    

\alpha=0.05 represent the significance level for the hypothesis test.    

t would represent the statistic

p_v represent the p value for the test

Hypothesis to test

We want to determine if the true mean is equal to 200, the system of hypothesis are :    

Null hypothesis:\mu = 200    

Alternative hypothesis:\mu = 200    

The statistic for this case is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)    

The statistic is given by:

t=\frac{201.77-200}{\frac{2.41}{\sqrt{10}}}=2.32    

The degrees of freedom are given by:

df=n-1=10-1=9  

The p value for this case is given by:

p_v =2*P(t_{(9)}>2.32)=0.0455    

For this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 200 F.

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Answer:

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Step-by-step explanation:

You can use f(15) = 40 to solve for C, then find f(0), the initial temperature.

  40 = f(15)

  40 = Ce^(-0.045·15) +14 = .50916C +14

  26 = .50916C

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Then f(0) is ...

  f(0) = 51.065·e^0 +14 = 65.065 ≈ 65

The initial temperature of the water was 65 degrees Fahrenheit.

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A quarter is approximately 0.07 inches thick. A dollar bill is approximately 4.3 × 10-3 inches thick.
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Step-by-step explanation:

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Last year, 46% of business owners gave a holiday gift to their employees. A survey of business owners indicated that 35% plan to
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Answer:

a) Number = 60 *0.35=21

b) Since is a left tailed test the p value would be:  

p_v =P(Z  

c) If we compare the p value obtained 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% the proportion of business owners providing holiday gifts had decreased from the 2008 level.  

We can use as smallest significance level 0.044 and we got the same conclusion.  

Step-by-step explanation:

Data given and notation  

n=60 represent the random sample taken

X represent the business owners plan to provide a holiday gift to their employees

\hat p=0.35 estimated proportion of business owners plan to provide a holiday gift to their employees

p_o=0.46 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)  

Part a

On this case w ejust need to multiply the value of th sample size by the proportion given like this:

Number = 60 *0.35=21

2) Part b

We need to conduct a hypothesis in order to test the claim that the proportion of business owners providing holiday gifts had decreased from the 2008 level:  

Null hypothesis:p\geq 0.46  

Alternative hypothesis:p < 0.46  

When we conduct a proportion test we need to use the z statisitc, 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.

Calculate the statistic  

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

z=\frac{0.35 -0.46}{\sqrt{\frac{0.46(1-0.46)}{60}}}=-1.710  

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 left tailed test the p value would be:  

p_v =P(Z  

Part c

If we compare the p value obtained 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% the proportion of business owners providing holiday gifts had decreased from the 2008 level.  

We can use as smallest significance level 0.044 and we got the same conclusion.  

8 0
1 year ago
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